PL9fwy3NUQKwaIc7KWc8buW344941GStQL/R6shw6IZsm8_postprocess.srt

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Last time, we talked about chi-square tests. And

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we mentioned that there are two objectives in this

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chapter. The first one is when to use chi-square

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tests for contingency tables. And the other

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objective is how to use chi-square tests for

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contingency tables. And we did one chi-square test

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for the difference between two proportions. In the

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null hypothesis, the two proportions are equal. I

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mean, proportion for population 1 equals

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population proportion 2 against the alternative

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here is two-sided test. Pi 1 does not equal pi 2.

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In this case, we can use either this statistic. So

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you may

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Z statistic, which is b1 minus b2 minus y1 minus

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y2 divided by b

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dash times 1 minus b dash multiplied by 1 over n1

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plus 1 over n2. This quantity under the square

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root, where b dash

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Or proportionally, where P dash equals X1 plus X2

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divided by N1 plus N2. Or,

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in this chapter, we are going to use chi-square

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statistic, which is given by this equation. Chi

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-square statistic is just sum of observed

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frequency, FO.

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minus expected frequency squared divided by

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expected frequency for all cells.

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Chi squared, this statistic is given by this

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equation. If there are two by two rows and

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columns, I mean there are two rows and two

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columns. So in this case, my table is two by two.

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In this case, you have only one degree of freedom.

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Always degrees of freedom equals number of rows

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minus one multiplied by number of columns minus

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one. So for two by two tables, there are two rows

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and two columns, so two minus one. times 2 minus

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1, so your degrees of freedom in this case is 1.

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Here the assumption is we assume that the expected

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frequency is at least 5, in order to use Chi

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-square statistic. Chi-square is always positive,

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I mean, Chi-square value is always greater than 0.

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It's one TLTS to the right one. We reject F0 if

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your chi-square statistic falls in the rejection

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region. That means we reject the null hypothesis

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if chi-square statistic greater than chi-square

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alpha. Alpha can be determined by using chi-square

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table. So we reject in this case F0, otherwise,

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sorry, we don't reject F0. So again, if the value

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of chi-square statistic falls in this rejection

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region, the yellow one, then we reject. Otherwise,

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if this value, I mean if the value of the

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statistic falls in non-rejection region, we don't

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reject the null hypothesis. So the same concept as

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we did in the previous chapters. If we go back to

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the previous example we had discussed before, when

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we are testing about gender and left and right

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handers, So hand preference either left or right.

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And the question is test to see whether hand

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preference and gender are related or not. In this

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case, your null hypothesis could be written as

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either X0.

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So the proportion of left-handers for female

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equals the proportion of males left-handers. So by

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one equals by two or H zero later we'll see that

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the two variables of interest are independent.

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Now, your B dash is

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given by X1 plus X2 divided by N1 plus N2. X1 is

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12, this 12, plus 24 divided by 300. That will

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give 12%. So let me just write this notation, B

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dash.

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equals 36 by 300, so that's 12%. So the expected

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frequency in this case for female, 0.12 times 120,

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because there are 120 females in the data you

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have, so that will give 14.4. So the expected

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frequency is 0.12 times 180, 120, I'm sorry,

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That will give 14.4. Similarly, for male to be

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left-handed is 0.12 times number of females in the

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sample, which is 180, and that will give 21.6.

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Now, since you compute the expected for the first

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cell, the second one direct is just the complement

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120. 120 is sample size for the Rome. I mean

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female total 120 minus 14.4 will give 105.6. Or 0

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.88 times 120 will give the same value. Here, the

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expected is 21.6, so the compliment is the, I'm

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sorry, the expected is just the compliment, which

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is 180 minus 21.6 will give 158.4. Or 0.88 is the

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compliment of that one multiplied by 180 will give

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the same value. So that's the one we had discussed

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before.

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On this result, you can determine the value of chi

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-square statistic by using this equation. Sum of F

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observed minus F expected squared divided by F

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expected for each cell. You have to compute the

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value of chi-square for each cell. In this case,

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the simplest case is just 2 by 2 table. So 12

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minus 14.4 squared divided by 14.4. Plus the

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second one 108 minus 105 squared divided by 105 up

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to the last one, you will get this result. Now my

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chi-square value is 0.7576.

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And in this case, if chi-square value is very

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small, I mean it's close to zero, then we don't

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reject the null hypothesis. Because the smallest

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value of chi-square is zero, and zero happens only

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if f observed is close to f expected. So here if

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you look carefully for the observed and expected

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frequencies, you can tell if you can reject or

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don't reject the number. Now the difference

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between these values looks small, so that's lead

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to small chi-square. So without doing the critical

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value, computer critical value, you can determine

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that we don't reject the null hypothesis. Because

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your chi-square value is very small. So we don't

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reject the null hypothesis. Or if you look

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carefully at the table, for the table we have

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here, for chi-square table. By the way, the

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smallest value of chi-square is 1.3. under 1

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degrees of freedom. So the smallest value 1.32. So

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if your chi-square value is greater than 1, it

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means maybe you reject or don't reject. It depends

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on v value and alpha you have or degrees of

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freedom. But in the worst scenario, if your chi

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-square is smaller than this value, it means you

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don't reject the null hypothesis. So generally

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speaking, if Chi-square is statistical. It's

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smaller than 1.32. 1.32 is a very small value.

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Then we don't reject. Then we don't reject x0.

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That's always, always true. Regardless of degrees

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of freedom and alpha. My chi-square is close to

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zero, or smaller than 1.32, because the minimum

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value of critical value is 1.32. Imagine that we

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are talking about alpha is 5%. So alpha is 5, so

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your critical value, the smallest one for 1

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degrees of freedom, is 3.84. So that's my

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smallest, if alpha

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Last time we mentioned that this value is just 1

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.96 squared. And that's only true, only true for 2

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by 2 table. That means this square is just Chi

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square 1. For this reason, we can test by one

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equal by two, by two methods, either this

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statistic or chi-square statistic. Both of them

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will give the same result. So let's go back to the

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question we have. My chi-square value is 0.77576.

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So that's your chi-square statistic. Again,

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degrees of freedom 1 to chi-square, the critical

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value is 3.841. So my decision is we don't reject

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the null hypothesis. My conclusion is there is not

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sufficient evidence that two proportions are

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different. So you don't have sufficient evidence

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in order to support that the two proportions are

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different at 5% level of significance. We stopped

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last time at this point. Now suppose we are

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testing The difference among more than two

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proportions. The same steps, we have to extend in

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this case chi-square. Your null hypothesis, by one

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equal by two, all the way up to by C. So in this

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case, there are C columns. C columns and

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two rows. So number of columns equals C, and there

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are only two rows. So pi 1 equals pi 2, all the

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way up to pi C. So null hypothesis for the columns

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we have. There are C columns. Again, it's the

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alternative, not all of the pi J are equal, and J

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equals 1 up to C. Now, the only difference here,

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the degrees of freedom.

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For 2 by c table,

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2 by c, degrees of freedom equals number

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of rows minus 1. There are two rows, so 2 minus 1

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times number of columns minus 1. 2 minus 1 is 1, c

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minus 1, 1 times c minus 1, c minus 1. So your

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degrees of freedom in this case is c minus 1.

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So that's the only difference. For two by two

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table, degrees of freedom is just one. If there

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are C columns and we have the same number of rows,

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degrees of freedom is C minus one. And we have the

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same chi squared statistic, the same equation I

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mean. And we have to extend also the overall

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proportion instead of x1 plus x2 divided by n1

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plus n2. It becomes x1 plus x2 plus x3 all the way

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up to xc because there are c columns divided by n1

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plus n2 all the way up to nc. So that's x over n.

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So similarly we can reject the null hypothesis if

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the value of chi-square statistic lies or falls in

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the rejection region.

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Other type of chi-square test is called chi-square

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test of independence. Generally speaking, most of

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the time there are more than two columns or more

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than two rows. Now, suppose we have contingency

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table that has R rows and C columns. And we are

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interested to test to see whether the two

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categorical variables are independent. That means

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there is no relationship between them. Against the

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alternative hypothesis, the two variables are

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dependent. That means there is a relationship

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between them. So test of independence.

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Null hypothesis is always the two variables, I

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mean, the two categorical variables are

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independent. So it's zero. Always x and y, for

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example, are independent.

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This means there is no difference between them. I

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mean, Y1 equals Y. Similarly, X and Y are

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independent. So there is no difference between the

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two populations of this notion. Against the

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alternative hypothesis, either X and Y, you may

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say that they are dependent.

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So that means there exists a relationship between

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them or They are related.

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So tests of independence for chi-square test to

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see whether or not the two variables are

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independent. So your null, two variables are

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independent against they are not independent. So

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similar to the chi-square test for equality of

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more than two proportions. So, in order to test to

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see if more than two proportions are equal, you

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cannot use this statistic. So, this statistic is

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no longer appropriate or valid for more than two

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proportions. In this case, you have to use chi

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-square test. So this statistic can be used only

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to test the difference between two proportions.

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But for more than two, you have to use chi-square

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test. So similar, chi-square test of independence

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is similar to chi-square test for equality of more

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than two proportions. But extend the concept. The

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previous one was two rows and C columns, so two by

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C. But here we extend the concept to contingency

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tables with R rows and C columns. So we have the

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case R by C. So that's in general, there are R

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rows and C columns. And the question is this C, if

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the two variables are independent or not. So in

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this case, you cannot use this statistic. So one

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more time, this statistic is valid only for two by

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two tables. So that means we can use z or chi

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-square to test if there is no difference between

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two population proportions. But for more than

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that, you have to use chi-square.

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Now still we have the same equation, Chi-square

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statistic is just sum F observed minus F expected

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quantity squared divided by F expected.

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In this case, Chi-square statistic for R by C case

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has degrees of freedom R minus 1 multiplied by C

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minus 1. In this case, each cell in the

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contingency table has expected frequency at least

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one instead of five. Now let's see how can we

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compute the expected cell frequency for each cell.

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The expected frequency is given by row total

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multiplied by colon total divided by n. So that's

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my new equation to determine I've expected it. So

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the expected value for each cell is given by Rho

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total multiplied by Kono, total divided by N.

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Also, this equation is true for the previous

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example. If you go back a little bit here, now the

249
00:20:16,650 --> 00:20:21,650
Expected for this cell was 40.4. Now let's see how

250
00:20:21,650 --> 00:20:25,470
can we compute the same value by using this

251
00:20:25,470 --> 00:20:30,250
equation. So it's equal to row total 120

252
00:20:30,250 --> 00:20:40,310
multiplied by column total 36 divided by 300.

253
00:20:43,580 --> 00:20:46,500
Now before we compute this value by using B dash

254
00:20:46,500 --> 00:20:50,900
first, 300 divided by, I'm sorry, 36 divided by

255
00:20:50,900 --> 00:20:58,520
300. So that's your B dash. Then we multiply this

256
00:20:58,520 --> 00:21:03,540
B dash by N, and this is your N. So it's similar

257
00:21:03,540 --> 00:21:08,540
equation. So either you use row total multiplied

258
00:21:08,540 --> 00:21:14,060
by column total. then divide by overall sample

259
00:21:14,060 --> 00:21:18,880
size you will get the same result by using the

260
00:21:18,880 --> 00:21:25,520
overall proportion 12% times 120 so each one will

261
00:21:25,520 --> 00:21:29,860
give the same answer so from now we are going to

262
00:21:29,860 --> 00:21:33,900
use this equation in order to compute the expected

263
00:21:33,900 --> 00:21:37,960
frequency for each cell so again expected

264
00:21:37,960 --> 00:21:42,920
frequency is rho total times Column total divided

265
00:21:42,920 --> 00:21:48,620
by N, N is the sample size. So row total it means

266
00:21:48,620 --> 00:21:52,220
sum of all frequencies in the row. Similarly

267
00:21:52,220 --> 00:21:56,160
column total is the sum of all frequencies in the

268
00:21:56,160 --> 00:22:00,180
column and N is over all sample size.

269
00:22:03,030 --> 00:22:06,630
Again, we reject the null hypothesis if your chi

270
00:22:06,630 --> 00:22:10,430
-square statistic greater than chi-square alpha.

271
00:22:10,590 --> 00:22:13,370
Otherwise, you don't reject it. And keep in mind,

272
00:22:14,270 --> 00:22:18,390
chi-square statistic has degrees of freedom R

273
00:22:18,390 --> 00:22:23,730
minus 1 times C minus 1. That's all for chi-square

274
00:22:23,730 --> 00:22:27,590
as test of independence. Any question?

275
00:22:31,220 --> 00:22:36,300
Here there is an example for applying chi-square

276
00:22:36,300 --> 00:22:42,200
test of independence. Meal plan selected

277
00:22:42,200 --> 00:22:46,700
by 200 students is shown in this table. So there

278
00:22:46,700 --> 00:22:50,960
are two variables of interest. The first one is

279
00:22:50,960 --> 00:22:56,230
number of meals per week. And there are three

280
00:22:56,230 --> 00:23:00,550
types of number of meals, either 20 meals per

281
00:23:00,550 --> 00:23:07,870
week, or 10 meals per week, or none. So that's, so

282
00:23:07,870 --> 00:23:12,150
number of meals is classified into three groups.

283
00:23:13,210 --> 00:23:17,650
So three columns, 20 per week, 10 per week, or

284
00:23:17,650 --> 00:23:23,270
none. Class standing, students are classified into

285
00:23:23,270 --> 00:23:28,860
four levels. A freshman, it means students like

286
00:23:28,860 --> 00:23:33,620
you, first year. Sophomore, it means second year.

287
00:23:34,440 --> 00:23:38,400
Junior, third level. Senior, fourth level. So that

288
00:23:38,400 --> 00:23:42,100
means first, second, third, and fourth level. And

289
00:23:42,100 --> 00:23:46,140
we have this number, these numbers for, I mean,

290
00:23:47,040 --> 00:23:53,660
there are 24 A freshman who have meals for 20 per

291
00:23:53,660 --> 00:23:59,880
week. So there are 24 freshmen have 20 meals per

292
00:23:59,880 --> 00:24:04,160
week. 22 sophomores, the same, 10 for junior and

293
00:24:04,160 --> 00:24:10,220
14 for senior. And the question is just to see if

294
00:24:10,220 --> 00:24:13,740
number of meals per week is independent of class

295
00:24:13,740 --> 00:24:17,270
standing. to see if there is a relationship

296
00:24:17,270 --> 00:24:21,890
between these two variables. In this case, there

297
00:24:21,890 --> 00:24:26,850
are four rows because the class standing is

298
00:24:26,850 --> 00:24:29,190
classified into four groups. So there are four

299
00:24:29,190 --> 00:24:34,230
rows and three columns. So this table actually is

300
00:24:34,230 --> 00:24:40,200
four by three. And there are twelve cells in this

301
00:24:40,200 --> 00:24:46,660
case. Now it takes time to compute the expected

302
00:24:46,660 --> 00:24:49,760
frequencies because in this case we have to

303
00:24:49,760 --> 00:24:55,120
compute the expected frequency for each cell. And

304
00:24:55,120 --> 00:25:01,320
we are going to use this formula for only six of

305
00:25:01,320 --> 00:25:06,260
them. I mean, we can apply this formula for only

306
00:25:06,260 --> 00:25:09,880
six of them. And the others can be computed by the

307
00:25:09,880 --> 00:25:14,300
complement by using either column total or row

308
00:25:14,300 --> 00:25:19,940
total. So because degrees of freedom is six, that

309
00:25:19,940 --> 00:25:23,880
means you may use this rule six times only. The

310
00:25:23,880 --> 00:25:28,420
others can be computed by using the complement. So

311
00:25:28,420 --> 00:25:34,070
here again, the hypothesis to be tested is, Mean

312
00:25:34,070 --> 00:25:36,550
plan and class standing are independent, that

313
00:25:36,550 --> 00:25:38,670
means there is no relationship between them.

314
00:25:39,150 --> 00:25:41,650
Against alternative hypothesis, mean plan and

315
00:25:41,650 --> 00:25:44,630
class standing are dependent, that means there

316
00:25:44,630 --> 00:25:49,950
exists significant relationship between them. Now

317
00:25:49,950 --> 00:25:54,390
let's see how can we compute the expected cell,

318
00:25:55,990 --> 00:26:00,470
the expected frequency for each cell. For example,

319
00:26:02,250 --> 00:26:07,790
The first observed frequency is 24. Now the

320
00:26:07,790 --> 00:26:15,990
expected should be 70 times 70 divided by 200. So

321
00:26:15,990 --> 00:26:25,050
for cell 11, the first cell. If expected, we can

322
00:26:25,050 --> 00:26:32,450
use this notation, 11. Means first row. First

323
00:26:32,450 --> 00:26:40,110
column. That should be 70. It is 70. Multiplied by

324
00:26:40,110 --> 00:26:43,990
column totals. Again, in this case, 70. Multiplied

325
00:26:43,990 --> 00:26:47,270
by 200. That will give 24.5.

326
00:26:50,150 --> 00:26:53,730
Similarly, for the second cell, for 32.

327
00:26:56,350 --> 00:27:00,090
70 times 88 divided by 200.

328
00:27:02,820 --> 00:27:12,620
So for F22, again it's 70 times 88 divided by 200,

329
00:27:12,800 --> 00:27:22,060
that will get 30.8. So 70 times 88, that will give

330
00:27:22,060 --> 00:27:32,780
30.8. F21, rule two first, one third. rho 1 second

331
00:27:32,780 --> 00:27:37,600
one the third one now either you can use the same

332
00:27:37,600 --> 00:27:44,320
equation which is 70 times 42 so you can use 70

333
00:27:44,320 --> 00:27:54,360
times 42 divided by 200 that will give 14.7 or

334
00:27:54,360 --> 00:27:59,000
it's just the complement which is 70 minus

335
00:28:03,390 --> 00:28:14,510
24.5 plus 30.8. So either use 70 multiplied by 40

336
00:28:14,510 --> 00:28:19,390
divided by 200 or just the complement, 70 minus.

337
00:28:20,800 --> 00:28:28,400
24.5 plus 30.8 will give the same value. So I just

338
00:28:28,400 --> 00:28:32,740
compute the expected cell for 1 and 2, and the

339
00:28:32,740 --> 00:28:36,120
third one is just the complement. Similarly, for

340
00:28:36,120 --> 00:28:42,560
the second row, I mean cell 21, then 22, and 23.

341
00:28:43,680 --> 00:28:47,940
By using the same method, he will get these two

342
00:28:47,940 --> 00:28:51,880
values, and the other one is the complement, which

343
00:28:51,880 --> 00:28:54,880
is 60 minus these, the sum of these two values,

344
00:28:55,300 --> 00:28:55,960
will give 12.

345
00:28:58,720 --> 00:29:01,920
Similarly, for the third cell, I'm sorry, the

346
00:29:01,920 --> 00:29:07,460
third row, for this value, For 10, it's 30 times

347
00:29:07,460 --> 00:29:12,660
70 divided by 200 will give this result. And the

348
00:29:12,660 --> 00:29:16,060
other one is just 30 multiplied by 88 divided by

349
00:29:16,060 --> 00:29:20,200
200. The other one is just the complement, 30

350
00:29:20,200 --> 00:29:25,180
minus the sum of these. Now, for the last column,

351
00:29:26,660 --> 00:29:35,220
either 70 multiplied by 70 divided by 200, or 70

352
00:29:35,220 --> 00:29:41,780
this 70 minus the sum of these. 70 this one equals

353
00:29:41,780 --> 00:29:51,740
70 minus the sum of 24 plus 21 plus 10. That will

354
00:29:51,740 --> 00:30:01,120
give 14. Now for the other expected cell, 88.

355
00:30:02,370 --> 00:30:05,530
minus the sum of these three expected frequencies.

356
00:30:07,290 --> 00:30:12,810
Now for the last one, last one is either by 42

357
00:30:12,810 --> 00:30:17,770
minus the sum of these three, or 40 minus the sum

358
00:30:17,770 --> 00:30:20,090
of 14 plus 6, 17.6.

359
00:30:22,810 --> 00:30:27,940
Or 40 multiplied by 42 divided by 400. So let's

360
00:30:27,940 --> 00:30:35,180
say we use that formula six times. For this

361
00:30:35,180 --> 00:30:39,100
reason, degrees of freedom is six. The other six

362
00:30:39,100 --> 00:30:46,480
are computed by the complement as we mentioned. So

363
00:30:46,480 --> 00:30:50,240
these are the expected frequencies. It takes time

364
00:30:50,240 --> 00:30:56,010
to compute these. But if you have only two by two

365
00:30:56,010 --> 00:31:01,170
table, it's easier. Now based on that, we can

366
00:31:01,170 --> 00:31:07,430
compute chi-square statistic value by using this

367
00:31:07,430 --> 00:31:12,390
equation for each cell. I mean, the first one, if

368
00:31:12,390 --> 00:31:14,370
you go back a little bit to the previous table,

369
00:31:15,150 --> 00:31:18,130
here, in order to compute chi-square,

370
00:31:22,640 --> 00:31:27,760
value, we have to use this equation, pi squared,

371
00:31:28,860 --> 00:31:36,080
sum F observed minus F expected squared, divided

372
00:31:36,080 --> 00:31:41,980
by F expected for all C's. So the first one is 24

373
00:31:41,980 --> 00:31:44,780
minus squared,

374
00:31:46,560 --> 00:31:55,350
24 plus. The second cell is 32 squared

375
00:31:55,350 --> 00:31:58,990
plus

376
00:31:58,990 --> 00:32:02,930
all the way up to the last cell, which is 10.

377
00:32:11,090 --> 00:32:14,430
So it takes time. But again, for two by two, it's

378
00:32:14,430 --> 00:32:18,890
straightforward. Anyway, now if you compare the

379
00:32:18,890 --> 00:32:23,650
expected and observed cells, you can have an idea

380
00:32:23,650 --> 00:32:25,650
either to reject or fail to reject without

381
00:32:25,650 --> 00:32:31,470
computing the value itself. Now, 24, 24.5. The

382
00:32:31,470 --> 00:32:32,430
difference is small.

383
00:32:35,730 --> 00:32:39,070
for about 7 and so on. So the difference between

384
00:32:39,070 --> 00:32:44,450
observed and expected looks small. In this case,

385
00:32:44,590 --> 00:32:50,530
chi-square value is close to zero. So it's 709.

386
00:32:51,190 --> 00:32:55,370
Now, without looking at the table we have, we have

387
00:32:55,370 --> 00:33:02,710
to don't reject. So we don't reject Because as we

388
00:33:02,710 --> 00:33:05,450
mentioned, the minimum k squared value is 1132.

389
00:33:06,350 --> 00:33:09,670
That's for one degrees of freedom and the alpha is

390
00:33:09,670 --> 00:33:14,390
25%. So

391
00:33:14,390 --> 00:33:19,250
I expect my decision is don't reject the null

392
00:33:19,250 --> 00:33:24,530
hypothesis. Now by looking at k squared 5% and

393
00:33:24,530 --> 00:33:28,870
degrees of freedom 6 by using k squared theorem.

394
00:33:30,200 --> 00:33:36,260
Now degrees of freedom 6. Now the minimum value of

395
00:33:36,260 --> 00:33:40,520
Chi-square is 7.84. I mean critical value. But

396
00:33:40,520 --> 00:33:48,290
under 5% is 12.59. So this value is 12.59. So

397
00:33:48,290 --> 00:33:54,470
critical value is 12.59. So my rejection region is

398
00:33:54,470 --> 00:33:59,890
above this value. Now, my chi-square value falls

399
00:33:59,890 --> 00:34:06,250
in the non-rejection regions. It's very small

400
00:34:06,250 --> 00:34:13,850
value. So chi-square statistic is 0.709.

401
00:34:14,230 --> 00:34:20,620
It's much smaller. Not even smaller than π²α, it's

402
00:34:20,620 --> 00:34:23,580
much smaller than this value, so it means we don't

403
00:34:23,580 --> 00:34:26,440
have sufficient evidence to support the

404
00:34:26,440 --> 00:34:32,010
alternative hypothesis. So my decision is, don't

405
00:34:32,010 --> 00:34:36,350
reject the null hypothesis. So conclusion, there

406
00:34:36,350 --> 00:34:41,150
is not sufficient evidence that Mealy Plan, which

407
00:34:41,150 --> 00:34:45,310
was classified into three groups, 20 per week or

408
00:34:45,310 --> 00:34:50,310
10 per week or none, and class standing. which is

409
00:34:50,310 --> 00:34:54,750
classified into four groups, freshman, sophomore,

410
00:34:55,010 --> 00:34:58,030
junior, and senior are related. So you don't have

411
00:34:58,030 --> 00:35:00,690
sufficient evidence that they are related. It

412
00:35:00,690 --> 00:35:05,630
means they are independent. So the two variables

413
00:35:05,630 --> 00:35:13,590
in this case are independent.

414
00:35:18,420 --> 00:35:21,520
It means there is no relationship between number

415
00:35:21,520 --> 00:35:25,000
of meals and class standing. It means the

416
00:35:25,000 --> 00:35:30,320
proportions are equal. So this means pi 1 equals

417
00:35:30,320 --> 00:35:34,560
pi 2 equals pi 3. So the three proportions are

418
00:35:34,560 --> 00:35:40,100
equal. Pi 1 for 20 meals per week is the same as

419
00:35:40,100 --> 00:35:46,960
10 or none according to class standing. Any

420
00:35:46,960 --> 00:35:52,600
question? I think it's straightforward test, maybe

421
00:35:52,600 --> 00:35:59,140
even easier than using a T statistic. And that's

422
00:35:59,140 --> 00:36:05,840
all for this chapter. Any questions? I will do

423
00:36:05,840 --> 00:36:12,360
some practice problems for chapter 11. These

424
00:36:12,360 --> 00:36:16,160
problems will be posted in the course website this

425
00:36:16,160 --> 00:36:19,220
week, sometime this week, maybe tomorrow or after

426
00:36:19,220 --> 00:36:22,840
tomorrow. So Monday or Tuesday I'm going to post

427
00:36:22,840 --> 00:36:27,280
the practice problems and solutions for chapter

428
00:36:27,280 --> 00:36:31,700
11. So let's do some of these problems.

429
00:36:40,160 --> 00:36:43,260
Let's do some of multiple choice problems.

430
00:36:55,000 --> 00:36:59,420
When testing for independence in contingency table

431
00:36:59,420 --> 00:37:03,840
with three rows and

432
00:37:03,840 --> 00:37:10,250
four columns. So there are three rows, four

433
00:37:10,250 --> 00:37:18,150
columns. There are degrees of freedom. So degrees

434
00:37:18,150 --> 00:37:23,310
of freedom. R minus one multiplied by C minus one.

435
00:37:24,090 --> 00:37:28,630
Two times three is six. So there are six degrees

436
00:37:28,630 --> 00:37:32,130
of freedom. Second question.

437
00:37:36,710 --> 00:37:43,150
If we wish to determine whether there is evidence

438
00:37:43,150 --> 00:37:46,890
that the proportion of items of interest is the

439
00:37:46,890 --> 00:37:51,510
same in group 1 as in group 2, the appropriate

440
00:37:51,510 --> 00:37:57,700
test to use is. So here we are testing Pi 1 equals

441
00:37:57,700 --> 00:38:01,040
Pi 2, so there are two populations.

442
00:38:02,480 --> 00:38:08,720
The answer is A. Z statistic, Z test, Chi squared,

443
00:38:09,740 --> 00:38:13,840
both A and B, neither A, neither of A nor B.

444
00:38:16,320 --> 00:38:19,540
Exactly, the answer is C because we can use either

445
00:38:19,540 --> 00:38:25,080
Z statistic or Chi squared. So Z or Chi. can be

446
00:38:25,080 --> 00:38:28,920
used for testing difference between two population

447
00:38:28,920 --> 00:38:34,360
proportions. And again, chi-square can be extended

448
00:38:34,360 --> 00:38:40,140
to use for more than two. So in this case, the

449
00:38:40,140 --> 00:38:43,220
correct answer is C, because we can use either Z

450
00:38:43,220 --> 00:38:52,090
or chi-square test. Next, in testing, hypothesis

451
00:38:52,090 --> 00:38:58,350
using chi-square test. The theoretical frequencies

452
00:38:58,350 --> 00:39:03,190
are based on null hypothesis, alternative, normal

453
00:39:03,190 --> 00:39:06,490
distribution, none of the above. Always when we

454
00:39:06,490 --> 00:39:10,450
are using chi-square test, we assume the null is

455
00:39:10,450 --> 00:39:14,630
true. So the theoretical frequencies are based on

456
00:39:14,630 --> 00:39:20,060
the null hypothesis. So always any statistic can

457
00:39:20,060 --> 00:39:25,300
be computed if we assume x0 is correct. So the

458
00:39:25,300 --> 00:39:26,400
correct answer is A.

459
00:39:34,060 --> 00:39:37,040
Let's look at table 11-2.

460
00:39:44,280 --> 00:39:49,000
Many companies use well-known celebrities as

461
00:39:49,000 --> 00:39:54,420
spokespersons in their TV advertisements. A study

462
00:39:54,420 --> 00:39:57,760
was conducted to determine whether brand awareness

463
00:39:57,760 --> 00:40:02,140
of female TV viewers and the gender of the

464
00:40:02,140 --> 00:40:05,860
spokesperson are independent. So there are two

465
00:40:05,860 --> 00:40:09,820
variables, whether a brand awareness of female TV

466
00:40:09,820 --> 00:40:13,740
and gender of the spokesperson are independent.

467
00:40:14,820 --> 00:40:19,540
Each and a sample of 300 female TV viewers was

468
00:40:19,540 --> 00:40:24,000
asked to identify a product advertised by a

469
00:40:24,000 --> 00:40:27,000
celebrity spokesperson, the gender of the

470
00:40:27,000 --> 00:40:32,280
spokesperson, and whether or not the viewer could

471
00:40:32,280 --> 00:40:36,460
identify the product was recorded. The number in

472
00:40:36,460 --> 00:40:40,080
each category are given below. Now, the questions

473
00:40:40,080 --> 00:40:45,520
are, number one, he asked about the calculated

474
00:40:45,520 --> 00:40:49,120
this statistic is. We have to find Chi-square

475
00:40:49,120 --> 00:40:54,020
statistic. It's two by two tables, easy one. So,

476
00:40:54,460 --> 00:40:59,460
for example, to find the F expected is,

477
00:41:00,420 --> 00:41:13,130
rho total is one over two. And one line here. And

478
00:41:13,130 --> 00:41:13,810
this 150.

479
00:41:16,430 --> 00:41:22,510
And also 150. So the expected frequency for the

480
00:41:22,510 --> 00:41:31,010
first one is 102 times 150 divided by 300.

481
00:41:35,680 --> 00:41:39,640
So the answer is 51.

482
00:41:42,880 --> 00:41:51,560
So the first expected is 51. The other one is just

483
00:41:51,560 --> 00:41:54,360
102 minus 51 is also 51.

484
00:41:57,320 --> 00:42:01,020
Now here is 99.

485
00:42:09,080 --> 00:42:15,180
So the second

486
00:42:15,180 --> 00:42:18,800
one are the expected frequencies. So my chi-square

487
00:42:18,800 --> 00:42:22,400
statistic is

488
00:42:22,400 --> 00:42:32,260
41 minus 51 squared divided by 51 plus 61 minus 51

489
00:42:32,260 --> 00:42:44,160
squared. 561 plus 109 minus 99 squared 99 plus 89

490
00:42:44,160 --> 00:42:47,040
minus 99 squared.

491
00:42:49,080 --> 00:42:53,140
That will give 5 point.

492
00:42:57,260 --> 00:43:01,760
So the answer is 5.9418.

493
00:43:03,410 --> 00:43:06,210
So simple calculation will give this result. Now,

494
00:43:06,450 --> 00:43:10,370
next one, referring to the same information we

495
00:43:10,370 --> 00:43:15,890
have at 5% level of significance, the critical

496
00:43:15,890 --> 00:43:18,510
value of that statistic. In this case, we are

497
00:43:18,510 --> 00:43:22,690
talking about 2 by 2 table, and alpha is 5. So

498
00:43:22,690 --> 00:43:28,130
your critical value is 3 point. So chi squared

499
00:43:28,130 --> 00:43:31,610
alpha, 5% and 1 degrees of freedom.

500
00:43:35,000 --> 00:43:39,220
This is the smallest value when alpha is 5%, so 3

501
00:43:39,220 --> 00:43:41,440
.8415.

502
00:43:46,160 --> 00:43:51,760
Again, degrees of freedom of this statistic are 1,

503
00:43:52,500 --> 00:43:53,800
2 by 2 is 1.

504
00:43:56,380 --> 00:44:01,620
Now at 5% level of significance, the conclusion is

505
00:44:01,620 --> 00:44:01,980
that

506
00:44:06,840 --> 00:44:16,380
In this case, we reject H0. And H0 says the two

507
00:44:16,380 --> 00:44:20,800
variables are independent. X and Y are

508
00:44:20,800 --> 00:44:25,860
independent. We reject that they are independent.

509
00:44:27,380 --> 00:44:33,200
That means they are dependent or related. So, A,

510
00:44:33,520 --> 00:44:36,680
brand awareness of female TV viewers and the

511
00:44:36,680 --> 00:44:41,380
gender of the spokesperson are independent. No,

512
00:44:41,580 --> 00:44:45,200
because we reject the null hypothesis. B, brand

513
00:44:45,200 --> 00:44:48,340
awareness of female TV viewers and the gender of

514
00:44:48,340 --> 00:44:53,140
spokesperson are not independent. Since we reject,

515
00:44:53,380 --> 00:44:58,330
then they are not. Because it's a complement. So,

516
00:44:58,430 --> 00:45:02,810
B is the correct answer. Now, C. A brand awareness

517
00:45:02,810 --> 00:45:05,450
of female TV viewers and the gender of the

518
00:45:05,450 --> 00:45:10,550
spokesperson are related. The same meaning. They

519
00:45:10,550 --> 00:45:15,470
are either, you say, not independent, related or

520
00:45:15,470 --> 00:45:15,950
dependent.

521
00:45:19,490 --> 00:45:24,930
Either is the same, so C is correct. D both B and

522
00:45:24,930 --> 00:45:28,970
C, so D is the correct answer. So again, if we

523
00:45:28,970 --> 00:45:31,650
reject the null hypothesis, it means the two

524
00:45:31,650 --> 00:45:36,990
variables either not independent or related or

525
00:45:36,990 --> 00:45:38,290
dependent.

526
00:45:40,550 --> 00:45:46,630
Any question? I will stop at this point. Next

527
00:45:46,630 --> 00:45:47,750
time, inshallah, we'll start.