Task 1
Confidence Interval
Given the weights of M and M's bags, we can calculate the sample mean (x̄) and sample standard deviation (s).

Using the given data, we find:

x̄ = 1.7466 (rounded to 4 decimal places)
s = 0.1075 (rounded to 4 decimal places)
n = 50 (sample size)

To construct a 95% confidence interval for the mean weight of a 1.69-ounce bag of M and M's, we first need to check the conditions for inference:

Randomness: We assume that the sample is representative of the population.
Normality: Since n > 30, the Central Limit Theorem states that the sampling distribution will be approximately normal.
Now, we can proceed with calculating the confidence interval. For a 95% confidence interval, we'll use the z-score for 97.5% (since we have a two-tailed test) which is approximately 1.96.

Margin of error (ME) = z * (s / sqrt(n))
ME = 1.96 * (0.1075 / sqrt(50))
ME ≈ 0.0301

Confidence Interval = (x̄ - ME, x̄ + ME)
CI = (1.7466 - 0.0301, 1.7466 + 0.0301)
CI ≈ (1.7165, 1.7767)

The 95% confidence interval represents that we are 95% confident that the true mean weight of a 1.69-ounce bag of M and M's is between 1.7165 and 1.7767 ounces.

Hypothesis Test
Now, we can perform a hypothesis test about the mean weight of a bag of M and M's at a 0.05 confidence level.

H0: µ = 1.69 (null hypothesis)
H1: µ ≠ 1.69 (alternative hypothesis)

Test statistic (z) = (x̄ - µ) / (s / sqrt(n))
z = (1.7466 - 1.69) / (0.1075 / sqrt(50))
z ≈ 3.71

At a 0.05 confidence level, our critical z-value for a two-tailed test is ±1.96. Since 3.71 > 1.96, we reject the null hypothesis (H0) and accept the alternative hypothesis (H1). This means that there is significant evidence to suggest that the mean weight of a 1.69-ounce bag of M and M's is not 1.69 ounces.

The conclusion of the hypothesis test is consistent with the confidence interval, as the interval does not contain the value 1.69, which is the mean weight stated in the null hypothesis.

Task 2
Hypothesis Test for Proportions
Let's choose two color categories: Yellow and Blue. We want to test if the proportions are equal.

H0: p_yellow = p_blue
H1: p_yellow ≠ p_blue

In our sample, we have:

n_yellow = 24
n_blue = 32
n_total = 107 (sum of all colors)

p̂_yellow = n_yellow / n_total = 24 / 107 ≈ 0.2243
p̂_blue = n_blue / n_total = 32 / 107 ≈ 0.2991

p̂ = (n_yellow + n_blue) / (2 * n_total) = (24 + 32) / (2 * 107) ≈ 0.2617

Test statistic (z) = (p̂_yellow - p̂_blue) / sqrt(p̂ * (1 - p̂) * (1 / n_yellow + 1 / n_blue))
z ≈ -1.97

At a 0.05 confidence level, our critical z-value for a two-tailed test is ±1.96. Since -1.97 < -1.96, we reject the null hypothesis (H0) and accept the alternative hypothesis (H1). This means that there is significant evidence to suggest that the proportions of yellow and blue M and M's are not equal.

Task 3
Chi-Square Goodness of Fit Test
We can perform a Chi-square goodness of fit test to determine if the observed color distribution matches the claimed distribution by Mars in 2008.

H0: The observed distribution matches the claimed distribution.
H1: The observed distribution does not match the claimed distribution.

Expected frequencies (E) for each color are calculated using the 2008 percentages and our sample size (n_total = 107):

E_brown = 0.13 * 107 ≈ 13.91
E_red = 0.13 *107 ≈ 13.91
E_yellow = 0.14 * 107 ≈ 14.98
E_green = 0.16 * 107 ≈ 17.12
E_orange = 0.20 * 107 ≈ 21.4
E_blue = 0.24 * 107 ≈ 25.68

Observed frequencies (O) are from the given data:

O_brown = 14
O_red = 11
O_yellow = 24
O_green = 8
O_orange = 18
O_blue = 32

Now, we calculate the Chi-square statistic (χ²):

χ² = Σ((O - E)² / E)
χ² ≈ (0.005)² / 13.91 + (-2.91)² / 13.91 + (9.02)² / 14.98 + (-9.12)² / 17.12 + (-3.4)² / 21.4 + (6.32)² / 25.68
χ² ≈ 15.03

Degrees of freedom (df) = number of categories - 1 = 6 - 1 = 5

Using a chi-square distribution table or calculator, we find the critical value for a 0.05 significance level and 5 degrees of freedom is approximately 11.07. Since 15.03 > 11.07, we reject the null hypothesis (H0) and accept the alternative hypothesis (H1). This means that there is significant evidence to suggest that the observed color distribution does not match the claimed distribution by Mars in 2008.

adjust the calculations for a mean of: 1.7572580645161 and a sample size (n) of 62, a standard deviation of: 0.06763376193566

here is the rubric for the project:
Task 1:

Weigh 2 bags of M and M’s separately and record the weight on the spreadsheet on Teams. Use all of your classmates’ data on the Teams spreadsheet to construct a 95% confidence interval for the mean weight of a 1.69 ounce bag of M and M’s. This will give us a sufficiently large sample of the weights. State what the confidence interval represents in the context of the question and perform all relevant statistical calculations showing all of the computation. You need to check all of the relevant conditions for inference.

Construct an appropriate hypothesis about the mean weight of a bag of M and M’s, and perform the necessary statistical analysis of the hypothesis, and verify all necessary conditions. Perform your testing at a 0.05 confidence level.

State your conclusion in the context of the problem, and discuss the connection between the hypothesis testing conclusion and the confidence interval.

Task 2:

Open both bags of M and M’s and count how many each color are present. Write your results in the spreadsheet on Teams under your name.

Pick two color categories and use your bags of M and M’s to construct a hypothesis about the proportion of M and M’s produced of each color based on your random sample of the population of all M and M’s produced. (Perform two separate tests.)

Test your hypothesis at a 0.05 confidence level and report the results showing all calculations in the context of the problem. State your conclusion in the context of the problem. (State, plan, do, conclude.)

Color How many

Brown

Red

Yellow

Green

Orange

Blue

Task 3:

Perform a Chi-square goodness of fit test based on the 2008 M and M production data provided.

Show all of the calculations leading to your conclusions. Make sure to verify the correct application of the Chi-square goodness of fit procedure.

Formulate and test an appropriate hypothesis on the claimed distribution of colors. The color distribution reported by Mars in 2008 was as follows:

Color Percentage

Brown 13%

Red 13%

Yellow 14%

Green 16%

Orange 20%

Blue 24%