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The dataset generation failed because of a cast error
Error code: DatasetGenerationCastError
Exception: DatasetGenerationCastError
Message: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 7 missing columns ({'question_property', 'story_id', 'graph_id', 'reasoning', 'formal_form', 'query_type', 'rung'})
This happened while the csv dataset builder was generating data using
hf://datasets/causal-nlp/CLadder/data/full_v1.csv (at revision 18d684e444ed07cc57ae0f3a91ee983a79b8119a)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2013, in _prepare_split_single
writer.write_table(table)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/arrow_writer.py", line 585, in write_table
pa_table = table_cast(pa_table, self._schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2302, in table_cast
return cast_table_to_schema(table, schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2256, in cast_table_to_schema
raise CastError(
datasets.table.CastError: Couldn't cast
id: int64
prompt: string
label: string
-- schema metadata --
pandas: '{"index_columns": [{"kind": "range", "name": null, "start": 0, "' + 586
to
{'id': Value(dtype='int64', id=None), 'prompt': Value(dtype='string', id=None), 'label': Value(dtype='string', id=None), 'reasoning': Value(dtype='string', id=None), 'rung': Value(dtype='int64', id=None), 'query_type': Value(dtype='string', id=None), 'graph_id': Value(dtype='string', id=None), 'story_id': Value(dtype='string', id=None), 'question_property': Value(dtype='string', id=None), 'formal_form': Value(dtype='string', id=None)}
because column names don't match
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1396, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1045, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1029, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1124, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1884, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2015, in _prepare_split_single
raise DatasetGenerationCastError.from_cast_error(
datasets.exceptions.DatasetGenerationCastError: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 7 missing columns ({'question_property', 'story_id', 'graph_id', 'reasoning', 'formal_form', 'query_type', 'rung'})
This happened while the csv dataset builder was generating data using
hf://datasets/causal-nlp/CLadder/data/full_v1.csv (at revision 18d684e444ed07cc57ae0f3a91ee983a79b8119a)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)Need help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
id
int64 | prompt
string | label
string | reasoning
string | rung
int64 | query_type
string | graph_id
string | story_id
string | question_property
string | formal_form
string |
|---|---|---|---|---|---|---|---|---|---|
4 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 8%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 54%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 41%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 86%. For husbands that don't set the alarm, the probability of alarm set by wife is 74%. For husbands that set the alarm, the probability of alarm set by wife is 24%. If we disregard the mediation effect through wife, would husband positively affect alarm clock? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.08
P(Y=1 | X=0, V2=1) = 0.54
P(Y=1 | X=1, V2=0) = 0.41
P(Y=1 | X=1, V2=1) = 0.86
P(V2=1 | X=0) = 0.74
P(V2=1 | X=1) = 0.24
0.74 * (0.86 - 0.41) + 0.24 * (0.54 - 0.08) = 0.32
0.32 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
7 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 26%. For husbands that set the alarm, the probability of ringing alarm is 76%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.26
P(Y=1 | X=1) = 0.76
0.76 - 0.26 = 0.50
0.50 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
8 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 77%. For husbands that don't set the alarm, the probability of ringing alarm is 26%. For husbands that set the alarm, the probability of ringing alarm is 76%. Is ringing alarm more likely than silent alarm overall? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.77
P(Y=1 | X=0) = 0.26
P(Y=1 | X=1) = 0.76
0.77*0.76 - 0.23*0.26 = 0.64
0.64 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
15 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 20%. For husbands that set the alarm, the probability of ringing alarm is 68%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.20
P(Y=1 | X=1) = 0.68
0.68 - 0.20 = 0.49
0.49 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
21 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 11%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 60%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 46%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 92%. For husbands that don't set the alarm, the probability of alarm set by wife is 61%. For husbands that set the alarm, the probability of alarm set by wife is 1%. Does husband positively affect alarm clock through wife? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.11
P(Y=1 | X=0, V2=1) = 0.60
P(Y=1 | X=1, V2=0) = 0.46
P(Y=1 | X=1, V2=1) = 0.92
P(V2=1 | X=0) = 0.61
P(V2=1 | X=1) = 0.01
0.01 * (0.60 - 0.11)+ 0.61 * (0.92 - 0.46)= -0.29
-0.29 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
22 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 11%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 60%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 46%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 92%. For husbands that don't set the alarm, the probability of alarm set by wife is 61%. For husbands that set the alarm, the probability of alarm set by wife is 1%. Does husband negatively affect alarm clock through wife? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.11
P(Y=1 | X=0, V2=1) = 0.60
P(Y=1 | X=1, V2=0) = 0.46
P(Y=1 | X=1, V2=1) = 0.92
P(V2=1 | X=0) = 0.61
P(V2=1 | X=1) = 0.01
0.01 * (0.60 - 0.11)+ 0.61 * (0.92 - 0.46)= -0.29
-0.29 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
27 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 3%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 50%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 45%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 94%. For husbands that don't set the alarm, the probability of alarm set by wife is 49%. For husbands that set the alarm, the probability of alarm set by wife is 53%. Does husband positively affect alarm clock through wife? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.03
P(Y=1 | X=0, V2=1) = 0.50
P(Y=1 | X=1, V2=0) = 0.45
P(Y=1 | X=1, V2=1) = 0.94
P(V2=1 | X=0) = 0.49
P(V2=1 | X=1) = 0.53
0.53 * (0.50 - 0.03)+ 0.49 * (0.94 - 0.45)= 0.02
0.02 > 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
28 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 88%. For husbands that don't set the alarm, the probability of ringing alarm is 26%. For husbands that set the alarm, the probability of ringing alarm is 71%. Is ringing alarm less likely than silent alarm overall? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.88
P(Y=1 | X=0) = 0.26
P(Y=1 | X=1) = 0.71
0.88*0.71 - 0.12*0.26 = 0.66
0.66 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
30 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look at how husband correlates with alarm clock case by case according to wife. Method 2: We look directly at how husband correlates with alarm clock in general. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
32 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 51%. For husbands that set the alarm, the probability of ringing alarm is 49%. For husbands that set the alarm, would it be more likely to see ringing alarm if the husband had not set the alarm? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.51
P(Y=1 | X=1) = 0.49
0.49 - 0.51 = -0.03
-0.03 < 0 | 3 | ett | mediation | alarm | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
43 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 42%. For husbands that set the alarm, the probability of ringing alarm is 66%. For husbands that set the alarm, would it be more likely to see ringing alarm if the husband had not set the alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.42
P(Y=1 | X=1) = 0.66
0.66 - 0.42 = 0.24
0.24 > 0 | 3 | ett | mediation | alarm | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
45 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 11%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 63%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 49%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 99%. For husbands that don't set the alarm, the probability of alarm set by wife is 59%. For husbands that set the alarm, the probability of alarm set by wife is 35%. Does husband positively affect alarm clock through wife? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.11
P(Y=1 | X=0, V2=1) = 0.63
P(Y=1 | X=1, V2=0) = 0.49
P(Y=1 | X=1, V2=1) = 0.99
P(V2=1 | X=0) = 0.59
P(V2=1 | X=1) = 0.35
0.35 * (0.63 - 0.11)+ 0.59 * (0.99 - 0.49)= -0.13
-0.13 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
60 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 44%. For husbands that set the alarm, the probability of ringing alarm is 75%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.44
P(Y=1 | X=1) = 0.75
0.75 - 0.44 = 0.30
0.30 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
62 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 8%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 51%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 49%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 92%. For husbands that don't set the alarm, the probability of alarm set by wife is 85%. For husbands that set the alarm, the probability of alarm set by wife is 60%. Does husband negatively affect alarm clock through wife? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.08
P(Y=1 | X=0, V2=1) = 0.51
P(Y=1 | X=1, V2=0) = 0.49
P(Y=1 | X=1, V2=1) = 0.92
P(V2=1 | X=0) = 0.85
P(V2=1 | X=1) = 0.60
0.60 * (0.51 - 0.08)+ 0.85 * (0.92 - 0.49)= -0.10
-0.10 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
66 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look directly at how husband correlates with alarm clock in general. Method 2: We look at this correlation case by case according to wife. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
69 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 9%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 55%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 42%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 86%. For husbands that don't set the alarm, the probability of alarm set by wife is 90%. For husbands that set the alarm, the probability of alarm set by wife is 30%. If we disregard the mediation effect through wife, would husband positively affect alarm clock? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.09
P(Y=1 | X=0, V2=1) = 0.55
P(Y=1 | X=1, V2=0) = 0.42
P(Y=1 | X=1, V2=1) = 0.86
P(V2=1 | X=0) = 0.90
P(V2=1 | X=1) = 0.30
0.90 * (0.86 - 0.42) + 0.30 * (0.55 - 0.09) = 0.31
0.31 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
72 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look at how husband correlates with alarm clock case by case according to wife. Method 2: We look directly at how husband correlates with alarm clock in general. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
75 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 71%. For husbands that don't set the alarm, the probability of ringing alarm is 39%. For husbands that set the alarm, the probability of ringing alarm is 62%. Is ringing alarm more likely than silent alarm overall? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.71
P(Y=1 | X=0) = 0.39
P(Y=1 | X=1) = 0.62
0.71*0.62 - 0.29*0.39 = 0.55
0.55 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
77 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 78%. For husbands that don't set the alarm, the probability of ringing alarm is 56%. For husbands that set the alarm, the probability of ringing alarm is 66%. Is ringing alarm less likely than silent alarm overall? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.78
P(Y=1 | X=0) = 0.56
P(Y=1 | X=1) = 0.66
0.78*0.66 - 0.22*0.56 = 0.64
0.64 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
80 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 81%. For husbands that don't set the alarm, the probability of ringing alarm is 21%. For husbands that set the alarm, the probability of ringing alarm is 86%. Is ringing alarm more likely than silent alarm overall? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.81
P(Y=1 | X=0) = 0.21
P(Y=1 | X=1) = 0.86
0.81*0.86 - 0.19*0.21 = 0.74
0.74 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
88 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 6%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 49%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 54%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 99%. For husbands that don't set the alarm, the probability of alarm set by wife is 94%. For husbands that set the alarm, the probability of alarm set by wife is 50%. If we disregard the mediation effect through wife, would husband positively affect alarm clock? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.06
P(Y=1 | X=0, V2=1) = 0.49
P(Y=1 | X=1, V2=0) = 0.54
P(Y=1 | X=1, V2=1) = 0.99
P(V2=1 | X=0) = 0.94
P(V2=1 | X=1) = 0.50
0.94 * (0.99 - 0.54) + 0.50 * (0.49 - 0.06) = 0.50
0.50 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
91 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look directly at how husband correlates with alarm clock in general. Method 2: We look at this correlation case by case according to wife. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
92 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 39%. For husbands that set the alarm, the probability of ringing alarm is 59%. Will alarm set by husband increase the chance of ringing alarm? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.39
P(Y=1 | X=1) = 0.59
0.59 - 0.39 = 0.20
0.20 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
94 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look at how husband correlates with alarm clock case by case according to wife. Method 2: We look directly at how husband correlates with alarm clock in general. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
99 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 38%. For husbands that set the alarm, the probability of ringing alarm is 60%. Will alarm set by husband increase the chance of ringing alarm? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.60
0.60 - 0.38 = 0.22
0.22 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
100 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 0%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 48%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 36%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 80%. For husbands that don't set the alarm, the probability of alarm set by wife is 79%. For husbands that set the alarm, the probability of alarm set by wife is 54%. Does husband negatively affect alarm clock through wife? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.00
P(Y=1 | X=0, V2=1) = 0.48
P(Y=1 | X=1, V2=0) = 0.36
P(Y=1 | X=1, V2=1) = 0.80
P(V2=1 | X=0) = 0.79
P(V2=1 | X=1) = 0.54
0.54 * (0.48 - 0.00)+ 0.79 * (0.80 - 0.36)= -0.12
-0.12 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
101 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 0%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 48%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 36%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 80%. For husbands that don't set the alarm, the probability of alarm set by wife is 79%. For husbands that set the alarm, the probability of alarm set by wife is 54%. If we disregard the mediation effect through wife, would husband positively affect alarm clock? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.00
P(Y=1 | X=0, V2=1) = 0.48
P(Y=1 | X=1, V2=0) = 0.36
P(Y=1 | X=1, V2=1) = 0.80
P(V2=1 | X=0) = 0.79
P(V2=1 | X=1) = 0.54
0.79 * (0.80 - 0.36) + 0.54 * (0.48 - 0.00) = 0.33
0.33 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
102 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. The overall probability of alarm set by husband is 70%. For husbands that don't set the alarm, the probability of ringing alarm is 38%. For husbands that set the alarm, the probability of ringing alarm is 60%. Is ringing alarm less likely than silent alarm overall? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.70
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.60
0.70*0.60 - 0.30*0.38 = 0.53
0.53 > 0 | 1 | marginal | mediation | alarm | easy | P(Y) |
113 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 8%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 50%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 39%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 86%. For husbands that don't set the alarm, the probability of alarm set by wife is 85%. For husbands that set the alarm, the probability of alarm set by wife is 29%. If we disregard the mediation effect through wife, would husband negatively affect alarm clock? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.08
P(Y=1 | X=0, V2=1) = 0.50
P(Y=1 | X=1, V2=0) = 0.39
P(Y=1 | X=1, V2=1) = 0.86
P(V2=1 | X=0) = 0.85
P(V2=1 | X=1) = 0.29
0.85 * (0.86 - 0.39) + 0.29 * (0.50 - 0.08) = 0.35
0.35 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
118 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 1%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 43%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 47%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 92%. For husbands that don't set the alarm, the probability of alarm set by wife is 50%. For husbands that set the alarm, the probability of alarm set by wife is 74%. Does husband positively affect alarm clock through wife? | yes | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.01
P(Y=1 | X=0, V2=1) = 0.43
P(Y=1 | X=1, V2=0) = 0.47
P(Y=1 | X=1, V2=1) = 0.92
P(V2=1 | X=0) = 0.50
P(V2=1 | X=1) = 0.74
0.74 * (0.43 - 0.01)+ 0.50 * (0.92 - 0.47)= 0.10
0.10 > 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
119 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 1%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 43%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 47%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 92%. For husbands that don't set the alarm, the probability of alarm set by wife is 50%. For husbands that set the alarm, the probability of alarm set by wife is 74%. If we disregard the mediation effect through wife, would husband negatively affect alarm clock? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=1, V2=0} - Y_{X=0, V2=0}]
\sum_{V2=v} P(V2=v|X=0)*[P(Y=1|X=1,V2=v) - P(Y=1|X=0, V2=v)]
P(Y=1 | X=0, V2=0) = 0.01
P(Y=1 | X=0, V2=1) = 0.43
P(Y=1 | X=1, V2=0) = 0.47
P(Y=1 | X=1, V2=1) = 0.92
P(V2=1 | X=0) = 0.50
P(V2=1 | X=1) = 0.74
0.50 * (0.92 - 0.47) + 0.74 * (0.43 - 0.01) = 0.47
0.47 > 0 | 3 | nde | mediation | alarm | easy | E[Y_{X=1, V2=0} - Y_{X=0, V2=0}] |
121 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 24%. For husbands that set the alarm, the probability of ringing alarm is 44%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.24
P(Y=1 | X=1) = 0.44
0.44 - 0.24 = 0.20
0.20 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
128 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 53%. For husbands that set the alarm, the probability of ringing alarm is 82%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.53
P(Y=1 | X=1) = 0.82
0.82 - 0.53 = 0.29
0.29 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
129 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 9%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 54%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 52%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 95%. For husbands that don't set the alarm, the probability of alarm set by wife is 98%. For husbands that set the alarm, the probability of alarm set by wife is 70%. Does husband positively affect alarm clock through wife? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.09
P(Y=1 | X=0, V2=1) = 0.54
P(Y=1 | X=1, V2=0) = 0.52
P(Y=1 | X=1, V2=1) = 0.95
P(V2=1 | X=0) = 0.98
P(V2=1 | X=1) = 0.70
0.70 * (0.54 - 0.09)+ 0.98 * (0.95 - 0.52)= -0.13
-0.13 < 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
133 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 36%. For husbands that set the alarm, the probability of ringing alarm is 55%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.36
P(Y=1 | X=1) = 0.55
0.55 - 0.36 = 0.18
0.18 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
138 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. Method 1: We look at how husband correlates with alarm clock case by case according to wife. Method 2: We look directly at how husband correlates with alarm clock in general. To understand how husband affects alarm clock, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | mediation | alarm | easy | [backdoor adjustment set for Y given X] |
143 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm, the probability of ringing alarm is 32%. For husbands that set the alarm, the probability of ringing alarm is 45%. Will alarm set by husband decrease the chance of ringing alarm? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.32
P(Y=1 | X=1) = 0.45
0.45 - 0.32 = 0.14
0.14 > 0 | 2 | ate | mediation | alarm | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
153 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: Husband has a direct effect on wife and alarm clock. Wife has a direct effect on alarm clock. For husbands that don't set the alarm and wives that don't set the alarm, the probability of ringing alarm is 3%. For husbands that don't set the alarm and wives that set the alarm, the probability of ringing alarm is 50%. For husbands that set the alarm and wives that don't set the alarm, the probability of ringing alarm is 50%. For husbands that set the alarm and wives that set the alarm, the probability of ringing alarm is 97%. For husbands that don't set the alarm, the probability of alarm set by wife is 63%. For husbands that set the alarm, the probability of alarm set by wife is 84%. Does husband negatively affect alarm clock through wife? | no | Let X = husband; V2 = wife; Y = alarm clock.
X->V2,X->Y,V2->Y
E[Y_{X=0, V2=1} - Y_{X=0, V2=0}]
\sum_{V2 = v} P(Y=1|X =0,V2 = v)*[P(V2 = v | X = 1) − P(V2 = v | X = 0)]
P(Y=1 | X=0, V2=0) = 0.03
P(Y=1 | X=0, V2=1) = 0.50
P(Y=1 | X=1, V2=0) = 0.50
P(Y=1 | X=1, V2=1) = 0.97
P(V2=1 | X=0) = 0.63
P(V2=1 | X=1) = 0.84
0.84 * (0.50 - 0.03)+ 0.63 * (0.97 - 0.50)= 0.10
0.10 > 0 | 3 | nie | mediation | alarm | easy | E[Y_{X=0, V2=1} - Y_{X=0, V2=0}] |
159 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 68%. The probability of not blowing out the candle and dark room is 12%. The probability of blowing out the candle and dark room is 51%. Is the chance of dark room smaller when observing blowing out the candle? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.68
P(Y=1, X=0=1) = 0.12
P(Y=1, X=1=1) = 0.51
0.51/0.68 - 0.12/0.32 = 0.38
0.38 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
160 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 38%. For people who blow out candles, the probability of dark room is 74%. Will blowing out the candle increase the chance of dark room? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.74
0.74 - 0.38 = 0.36
0.36 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
162 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 38%. For people who blow out candles, the probability of dark room is 74%. For people who blow out candles, would it be more likely to see dark room if the person in the room had not blown out the candle? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.74
0.74 - 0.38 = 0.36
0.36 > 0 | 3 | ett | fork | candle | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
166 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 44%. For people who blow out candles, the probability of dark room is 75%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.44
P(Y=1 | X=1) = 0.75
0.75 - 0.44 = 0.31
0.31 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
173 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 56%. For people not blowing out candles, the probability of dark room is 55%. For people who blow out candles, the probability of dark room is 89%. Is dark room less likely than bright room overall? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.56
P(Y=1 | X=0) = 0.55
P(Y=1 | X=1) = 0.89
0.56*0.89 - 0.44*0.55 = 0.74
0.74 > 0 | 1 | marginal | fork | candle | easy | P(Y) |
175 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 39%. For people who blow out candles, the probability of dark room is 72%. For people who blow out candles, would it be less likely to see dark room if the person in the room had not blown out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.39
P(Y=1 | X=1) = 0.72
0.72 - 0.39 = 0.33
0.33 > 0 | 3 | ett | fork | candle | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
176 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 43%. The probability of not blowing out the candle and dark room is 22%. The probability of blowing out the candle and dark room is 31%. Is the chance of dark room larger when observing blowing out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.43
P(Y=1, X=0=1) = 0.22
P(Y=1, X=1=1) = 0.31
0.31/0.43 - 0.22/0.57 = 0.33
0.33 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
178 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look at how the man in the room correlates with room case by case according to the candle. Method 2: We look directly at how the man in the room correlates with room in general. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
180 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 31%. For people who blow out candles, the probability of dark room is 66%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.31
P(Y=1 | X=1) = 0.66
0.66 - 0.31 = 0.34
0.34 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
184 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look directly at how the man in the room correlates with room in general. Method 2: We look at this correlation case by case according to the candle. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
194 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 31%. The probability of not blowing out the candle and dark room is 21%. The probability of blowing out the candle and dark room is 19%. Is the chance of dark room larger when observing blowing out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.31
P(Y=1, X=0=1) = 0.21
P(Y=1, X=1=1) = 0.19
0.19/0.31 - 0.21/0.69 = 0.31
0.31 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
197 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 38%. For people who blow out candles, the probability of dark room is 72%. For people who blow out candles, would it be more likely to see dark room if the person in the room had not blown out the candle? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.72
0.72 - 0.38 = 0.34
0.34 > 0 | 3 | ett | fork | candle | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
199 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 29%. The probability of not blowing out the candle and dark room is 27%. The probability of blowing out the candle and dark room is 21%. Is the chance of dark room smaller when observing blowing out the candle? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.29
P(Y=1, X=0=1) = 0.27
P(Y=1, X=1=1) = 0.21
0.21/0.29 - 0.27/0.71 = 0.34
0.34 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
201 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 48%. For people who blow out candles, the probability of dark room is 79%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.48
P(Y=1 | X=1) = 0.79
0.79 - 0.48 = 0.32
0.32 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
203 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 50%. The probability of not blowing out the candle and dark room is 24%. The probability of blowing out the candle and dark room is 40%. Is the chance of dark room larger when observing blowing out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.50
P(Y=1, X=0=1) = 0.24
P(Y=1, X=1=1) = 0.40
0.40/0.50 - 0.24/0.50 = 0.32
0.32 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
204 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look directly at how the man in the room correlates with room in general. Method 2: We look at this correlation case by case according to the candle. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
207 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 51%. For people who blow out candles, the probability of dark room is 93%. For people who blow out candles, would it be less likely to see dark room if the person in the room had not blown out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.51
P(Y=1 | X=1) = 0.93
0.93 - 0.51 = 0.42
0.42 > 0 | 3 | ett | fork | candle | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
213 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 38%. For people who blow out candles, the probability of dark room is 73%. Will blowing out the candle increase the chance of dark room? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.38
P(Y=1 | X=1) = 0.73
0.73 - 0.38 = 0.35
0.35 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
216 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 30%. For people who blow out candles, the probability of dark room is 63%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.30
P(Y=1 | X=1) = 0.63
0.63 - 0.30 = 0.33
0.33 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
218 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 65%. The probability of not blowing out the candle and dark room is 10%. The probability of blowing out the candle and dark room is 41%. Is the chance of dark room smaller when observing blowing out the candle? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.65
P(Y=1, X=0=1) = 0.10
P(Y=1, X=1=1) = 0.41
0.41/0.65 - 0.10/0.35 = 0.33
0.33 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
224 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 23%. For people not blowing out candles, the probability of dark room is 23%. For people who blow out candles, the probability of dark room is 58%. Is dark room more likely than bright room overall? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.23
P(Y=1 | X=0) = 0.23
P(Y=1 | X=1) = 0.58
0.23*0.58 - 0.77*0.23 = 0.31
0.31 > 0 | 1 | marginal | fork | candle | easy | P(Y) |
229 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 43%. For people who blow out candles, the probability of dark room is 74%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.43
P(Y=1 | X=1) = 0.74
0.74 - 0.43 = 0.31
0.31 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
230 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 53%. For people not blowing out candles, the probability of dark room is 43%. For people who blow out candles, the probability of dark room is 74%. Is dark room more likely than bright room overall? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.53
P(Y=1 | X=0) = 0.43
P(Y=1 | X=1) = 0.74
0.53*0.74 - 0.47*0.43 = 0.59
0.59 > 0 | 1 | marginal | fork | candle | easy | P(Y) |
232 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look at how the man in the room correlates with room case by case according to the candle. Method 2: We look directly at how the man in the room correlates with room in general. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
236 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 41%. For people who blow out candles, the probability of dark room is 74%. Will blowing out the candle increase the chance of dark room? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.41
P(Y=1 | X=1) = 0.74
0.74 - 0.41 = 0.32
0.32 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
237 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 41%. For people who blow out candles, the probability of dark room is 74%. For people who blow out candles, would it be less likely to see dark room if the person in the room had not blown out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
P(Y=1|X=1) - P(Y=1|X=0)
P(Y=1 | X=0) = 0.41
P(Y=1 | X=1) = 0.74
0.74 - 0.41 = 0.32
0.32 > 0 | 3 | ett | fork | candle | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
240 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 43%. For people not blowing out candles, the probability of dark room is 48%. For people who blow out candles, the probability of dark room is 80%. Is dark room more likely than bright room overall? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.43
P(Y=1 | X=0) = 0.48
P(Y=1 | X=1) = 0.80
0.43*0.80 - 0.57*0.48 = 0.61
0.61 > 0 | 1 | marginal | fork | candle | easy | P(Y) |
241 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 41%. For people who blow out candles, the probability of dark room is 74%. Will blowing out the candle increase the chance of dark room? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.41
P(Y=1 | X=1) = 0.74
0.74 - 0.41 = 0.33
0.33 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
247 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look at how the man in the room correlates with room case by case according to the candle. Method 2: We look directly at how the man in the room correlates with room in general. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
249 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 56%. For people who blow out candles, the probability of dark room is 90%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.56
P(Y=1 | X=1) = 0.90
0.90 - 0.56 = 0.34
0.34 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
254 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look at how the man in the room correlates with room case by case according to the candle. Method 2: We look directly at how the man in the room correlates with room in general. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
255 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look directly at how the man in the room correlates with room in general. Method 2: We look at this correlation case by case according to the candle. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
265 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. The overall probability of blowing out the candle is 26%. The probability of not blowing out the candle and dark room is 41%. The probability of blowing out the candle and dark room is 22%. Is the chance of dark room larger when observing blowing out the candle? | yes | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.26
P(Y=1, X=0=1) = 0.41
P(Y=1, X=1=1) = 0.22
0.22/0.26 - 0.41/0.74 = 0.31
0.31 > 0 | 1 | correlation | fork | candle | easy | P(Y | X) |
269 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. For people not blowing out candles, the probability of dark room is 33%. For people who blow out candles, the probability of dark room is 68%. Will blowing out the candle decrease the chance of dark room? | no | Let V2 = the candle; X = the man in the room; Y = room.
X->Y,V2->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
P(Y|X)
P(Y=1 | X=0) = 0.33
P(Y=1 | X=1) = 0.68
0.68 - 0.33 = 0.35
0.35 > 0 | 2 | ate | fork | candle | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
273 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: The man in the room has a direct effect on room. The candle has a direct effect on room. Method 1: We look directly at how the man in the room correlates with room in general. Method 2: We look at this correlation case by case according to the candle. To understand how the man in the room affects room, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | fork | candle | easy | [backdoor adjustment set for Y given X] |
275 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 86%. For managers who don't sign termination letters, the probability of employee being fired is 31%. For managers who sign termination letters, the probability of employee being fired is 70%. Is employee being fired more likely than employee not being fired overall? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.86
P(Y=1 | X=0) = 0.31
P(Y=1 | X=1) = 0.70
0.86*0.70 - 0.14*0.31 = 0.65
0.65 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
276 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 86%. The probability of manager not signing the termination letter and employee being fired is 4%. The probability of manager signing the termination letter and employee being fired is 60%. Is the chance of employee being fired larger when observing manager signing the termination letter? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.86
P(Y=1, X=0=1) = 0.04
P(Y=1, X=1=1) = 0.60
0.60/0.86 - 0.04/0.14 = 0.39
0.39 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
278 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 9%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 54%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 46%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 90%. For managers who don't sign termination letters, the probability of director signing the termination letter is 17%. For managers who sign termination letters, the probability of director signing the termination letter is 27%. For managers who sign termination letters, would it be less likely to see employee being fired if the manager had signed the termination letter? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.09
P(Y=1 | X=0, V3=1) = 0.54
P(Y=1 | X=1, V3=0) = 0.46
P(Y=1 | X=1, V3=1) = 0.90
P(V3=1 | X=0) = 0.17
P(V3=1 | X=1) = 0.27
0.27 * (0.90 - 0.46) + 0.17 * (0.54 - 0.09) = 0.37
0.37 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
279 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 40%. For managers who don't sign termination letters, the probability of employee being fired is 16%. For managers who sign termination letters, the probability of employee being fired is 58%. Is employee being fired less likely than employee not being fired overall? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.40
P(Y=1 | X=0) = 0.16
P(Y=1 | X=1) = 0.58
0.40*0.58 - 0.60*0.16 = 0.33
0.33 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
281 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 40%. The probability of manager not signing the termination letter and employee being fired is 10%. The probability of manager signing the termination letter and employee being fired is 23%. Is the chance of employee being fired smaller when observing manager signing the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.40
P(Y=1, X=0=1) = 0.10
P(Y=1, X=1=1) = 0.23
0.23/0.40 - 0.10/0.60 = 0.42
0.42 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
282 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. Method 1: We look directly at how manager correlates with employee in general. Method 2: We look at this correlation case by case according to director. To understand how manager affects employee, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | diamondcut | firing_employee | easy | [backdoor adjustment set for Y given X] |
284 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 13%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 51%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 46%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 83%. For managers who don't sign termination letters, the probability of director signing the termination letter is 53%. For managers who sign termination letters, the probability of director signing the termination letter is 60%. For managers who sign termination letters, would it be more likely to see employee being fired if the manager had signed the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.13
P(Y=1 | X=0, V3=1) = 0.51
P(Y=1 | X=1, V3=0) = 0.46
P(Y=1 | X=1, V3=1) = 0.83
P(V3=1 | X=0) = 0.53
P(V3=1 | X=1) = 0.60
0.60 * (0.83 - 0.46) + 0.53 * (0.51 - 0.13) = 0.33
0.33 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
286 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 23%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 57%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 43%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 79%. The overall probability of CEO's decision to fire the employee is 17%. Will manager signing the termination letter increase the chance of employee being fired? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.23
P(Y=1 | V1=0, X=1) = 0.57
P(Y=1 | V1=1, X=0) = 0.43
P(Y=1 | V1=1, X=1) = 0.79
P(V1=1) = 0.17
0.83 * (0.57 - 0.23) 0.17 * (0.79 - 0.43) = 0.35
0.35 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
287 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 23%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 57%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 43%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 79%. The overall probability of CEO's decision to fire the employee is 17%. Will manager signing the termination letter decrease the chance of employee being fired? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.23
P(Y=1 | V1=0, X=1) = 0.57
P(Y=1 | V1=1, X=0) = 0.43
P(Y=1 | V1=1, X=1) = 0.79
P(V1=1) = 0.17
0.83 * (0.57 - 0.23) 0.17 * (0.79 - 0.43) = 0.35
0.35 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
289 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 32%. The probability of manager not signing the termination letter and employee being fired is 16%. The probability of manager signing the termination letter and employee being fired is 21%. Is the chance of employee being fired smaller when observing manager signing the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.32
P(Y=1, X=0=1) = 0.16
P(Y=1, X=1=1) = 0.21
0.21/0.32 - 0.16/0.68 = 0.42
0.42 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
290 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 49%. The probability of manager not signing the termination letter and employee being fired is 14%. The probability of manager signing the termination letter and employee being fired is 37%. Is the chance of employee being fired smaller when observing manager signing the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.49
P(Y=1, X=0=1) = 0.14
P(Y=1, X=1=1) = 0.37
0.37/0.49 - 0.14/0.51 = 0.48
0.48 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
291 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 16%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 52%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 37%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 71%. The overall probability of CEO's decision to fire the employee is 13%. Will manager signing the termination letter increase the chance of employee being fired? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.16
P(Y=1 | V1=0, X=1) = 0.52
P(Y=1 | V1=1, X=0) = 0.37
P(Y=1 | V1=1, X=1) = 0.71
P(V1=1) = 0.13
0.87 * (0.52 - 0.16) 0.13 * (0.71 - 0.37) = 0.36
0.36 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
292 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 10%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 52%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 47%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 84%. For managers who don't sign termination letters, the probability of director signing the termination letter is 19%. For managers who sign termination letters, the probability of director signing the termination letter is 26%. For managers who sign termination letters, would it be more likely to see employee being fired if the manager had signed the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.10
P(Y=1 | X=0, V3=1) = 0.52
P(Y=1 | X=1, V3=0) = 0.47
P(Y=1 | X=1, V3=1) = 0.84
P(V3=1 | X=0) = 0.19
P(V3=1 | X=1) = 0.26
0.26 * (0.84 - 0.47) + 0.19 * (0.52 - 0.10) = 0.36
0.36 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
293 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 30%. For managers who don't sign termination letters, the probability of employee being fired is 18%. For managers who sign termination letters, the probability of employee being fired is 57%. Is employee being fired more likely than employee not being fired overall? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.30
P(Y=1 | X=0) = 0.18
P(Y=1 | X=1) = 0.57
0.30*0.57 - 0.70*0.18 = 0.30
0.30 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
294 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 30%. For managers who don't sign termination letters, the probability of employee being fired is 18%. For managers who sign termination letters, the probability of employee being fired is 57%. Is employee being fired less likely than employee not being fired overall? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.30
P(Y=1 | X=0) = 0.18
P(Y=1 | X=1) = 0.57
0.30*0.57 - 0.70*0.18 = 0.30
0.30 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
295 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 30%. The probability of manager not signing the termination letter and employee being fired is 13%. The probability of manager signing the termination letter and employee being fired is 17%. Is the chance of employee being fired smaller when observing manager signing the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.30
P(Y=1, X=0=1) = 0.13
P(Y=1, X=1=1) = 0.17
0.17/0.30 - 0.13/0.70 = 0.39
0.39 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
297 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 6%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 44%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 48%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 89%. For managers who don't sign termination letters, the probability of director signing the termination letter is 33%. For managers who sign termination letters, the probability of director signing the termination letter is 44%. For managers who sign termination letters, would it be more likely to see employee being fired if the manager had signed the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.06
P(Y=1 | X=0, V3=1) = 0.44
P(Y=1 | X=1, V3=0) = 0.48
P(Y=1 | X=1, V3=1) = 0.89
P(V3=1 | X=0) = 0.33
P(V3=1 | X=1) = 0.44
0.44 * (0.89 - 0.48) + 0.33 * (0.44 - 0.06) = 0.44
0.44 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
300 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 12%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 45%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 47%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 88%. The overall probability of CEO's decision to fire the employee is 30%. Will manager signing the termination letter decrease the chance of employee being fired? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.12
P(Y=1 | V1=0, X=1) = 0.45
P(Y=1 | V1=1, X=0) = 0.47
P(Y=1 | V1=1, X=1) = 0.88
P(V1=1) = 0.30
0.70 * (0.45 - 0.12) 0.30 * (0.88 - 0.47) = 0.36
0.36 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
302 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. Method 1: We look directly at how manager correlates with employee in general. Method 2: We look at this correlation case by case according to director. To understand how manager affects employee, is it more correct to use the Method 1 than Method 2? | no | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | diamondcut | firing_employee | easy | [backdoor adjustment set for Y given X] |
304 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 31%. The probability of manager not signing the termination letter and employee being fired is 21%. The probability of manager signing the termination letter and employee being fired is 24%. Is the chance of employee being fired larger when observing manager signing the termination letter? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y | X)
P(X = 1, Y = 1)/P(X = 1) - P(X = 0, Y = 1)/P(X = 0)
P(X=1=1) = 0.31
P(Y=1, X=0=1) = 0.21
P(Y=1, X=1=1) = 0.24
0.24/0.31 - 0.21/0.69 = 0.47
0.47 > 0 | 1 | correlation | diamondcut | firing_employee | easy | P(Y | X) |
307 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 22%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 61%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 44%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 82%. The overall probability of CEO's decision to fire the employee is 22%. Will manager signing the termination letter increase the chance of employee being fired? | yes | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.22
P(Y=1 | V1=0, X=1) = 0.61
P(Y=1 | V1=1, X=0) = 0.44
P(Y=1 | V1=1, X=1) = 0.82
P(V1=1) = 0.22
0.78 * (0.61 - 0.22) 0.22 * (0.82 - 0.44) = 0.39
0.39 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
308 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 17%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 52%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 56%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 90%. For managers who don't sign termination letters, the probability of director signing the termination letter is 23%. For managers who sign termination letters, the probability of director signing the termination letter is 38%. For managers who sign termination letters, would it be more likely to see employee being fired if the manager had signed the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.17
P(Y=1 | X=0, V3=1) = 0.52
P(Y=1 | X=1, V3=0) = 0.56
P(Y=1 | X=1, V3=1) = 0.90
P(V3=1 | X=0) = 0.23
P(V3=1 | X=1) = 0.38
0.38 * (0.90 - 0.56) + 0.23 * (0.52 - 0.17) = 0.39
0.39 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
309 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 40%. For managers who don't sign termination letters, the probability of employee being fired is 25%. For managers who sign termination letters, the probability of employee being fired is 69%. Is employee being fired more likely than employee not being fired overall? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.40
P(Y=1 | X=0) = 0.25
P(Y=1 | X=1) = 0.69
0.40*0.69 - 0.60*0.25 = 0.42
0.42 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
312 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 43%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 81%. For CEOs who fire employees and managers who don't sign termination letters, the probability of employee being fired is 63%. For CEOs who fire employees and managers who sign termination letters, the probability of employee being fired is 98%. The overall probability of CEO's decision to fire the employee is 26%. Will manager signing the termination letter decrease the chance of employee being fired? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y | do(X = 1)] - E[Y | do(X = 0)]
\sum_{V1=v} P(V1=v)*[P(Y=1|V1=v,X=1) - P(Y=1|V1=v, X=0)]
P(Y=1 | V1=0, X=0) = 0.43
P(Y=1 | V1=0, X=1) = 0.81
P(Y=1 | V1=1, X=0) = 0.63
P(Y=1 | V1=1, X=1) = 0.98
P(V1=1) = 0.26
0.74 * (0.81 - 0.43) 0.26 * (0.98 - 0.63) = 0.38
0.38 > 0 | 2 | ate | diamondcut | firing_employee | easy | E[Y | do(X = 1)] - E[Y | do(X = 0)] |
313 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. For managers who don't sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 21%. For managers who don't sign termination letters and directors who sign termination letters, the probability of employee being fired is 64%. For managers who sign termination letters and directors who don't sign termination letters, the probability of employee being fired is 63%. For managers who sign termination letters and directors who sign termination letters, the probability of employee being fired is 99%. For managers who don't sign termination letters, the probability of director signing the termination letter is 60%. For managers who sign termination letters, the probability of director signing the termination letter is 64%. For managers who sign termination letters, would it be more likely to see employee being fired if the manager had signed the termination letter? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
E[Y_{X = 1} - Y_{X = 0} | X = 1]
\sum_{V3=v} P(V3=v|X=1)*[P(Y=1|X=1,V3=v) - P(Y=1|X=0,V3=v)]
P(Y=1 | X=0, V3=0) = 0.21
P(Y=1 | X=0, V3=1) = 0.64
P(Y=1 | X=1, V3=0) = 0.63
P(Y=1 | X=1, V3=1) = 0.99
P(V3=1 | X=0) = 0.60
P(V3=1 | X=1) = 0.64
0.64 * (0.99 - 0.63) + 0.60 * (0.64 - 0.21) = 0.38
0.38 > 0 | 3 | ett | diamondcut | firing_employee | easy | E[Y_{X = 1} - Y_{X = 0} | X = 1] |
315 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. Method 1: We look at how manager correlates with employee case by case according to director. Method 2: We look directly at how manager correlates with employee in general. To understand how manager affects employee, is it more correct to use the Method 1 than Method 2? | yes | nan
nan
nan
nan
nan
nan
nan | 2 | backadj | diamondcut | firing_employee | easy | [backdoor adjustment set for Y given X] |
318 | Imagine a self-contained, hypothetical world with only the following conditions, and without any unmentioned factors or causal relationships: CEO has a direct effect on director and manager. Manager has a direct effect on employee. Director has a direct effect on employee. The overall probability of manager signing the termination letter is 61%. For managers who don't sign termination letters, the probability of employee being fired is 31%. For managers who sign termination letters, the probability of employee being fired is 67%. Is employee being fired less likely than employee not being fired overall? | no | Let V1 = CEO; V3 = director; X = manager; Y = employee.
V1->V3,V1->X,X->Y,V3->Y
P(Y)
P(Y | X=1)*P(X=1) + P(Y | X=0)*P(X=0)
P(X=1) = 0.61
P(Y=1 | X=0) = 0.31
P(Y=1 | X=1) = 0.67
0.61*0.67 - 0.39*0.31 = 0.53
0.53 > 0 | 1 | marginal | diamondcut | firing_employee | easy | P(Y) |
End of preview.