AP Statistics 3.15 Chi-Square Conclusions Overview

[Maximum number: 1]

Questions 6-8 refer to the following situation: A computer manufacturer sets up three locations to provide technical support for its customers. Logs are kept, noting whether or not calls about problems are solved successfully. Data from a sample of 1,000 calls are summarized in the following table:

Table for Question 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence question 1 — AP Statistics

Assuming there is no association between location and whether or not a problem is resolved successfully, what is the expected number of successful calls (problem solved) from location 1?

$\frac{(325)(450)}{700}$

$\frac{(450)(700)}{1,000}$

$\frac{(325)(700)}{450}$

$\frac{(325)(700)}{1,000}$

．Three months from now，all of the parents／guardians of elementary school students in a certain large school

district will vote either yes or no on a new electronic tablet program．This program will allow every elementary

school student within the school district to be issued a tablet．In a recent survey of a random sample of 300

parents／guardians of elementary school students， 146 out of 300 （48．7 percent）indicated they would vote yes in

favor of such a program．

Mr．Robinson，the technology expert at the school district，is in favor of the elementary students being issued a

tablet．He develops a pamphlet showing the educational benefits of students having their own tablet．He did this to

increase the proportion of parents／guardians voting yes．To investigate the effectiveness of the pamphlet，Mr．

Robinson distributed the pamphlet to the same 300 parents／guardians from the original sample．After reading the

pamphlet，each parent／guardian was asked again whether each elementary student should be issued a tablet．

Parents／guardians could change their answer or keep the same answer from the original survey．The results of the

parent／guardian replies are shown in the table．

Table for Question 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence question 2 — AP Statistics

(a)

（c）（i）Assume the conditions for McNemar＇s test have been met．Calculate the test statistic using the sample data

from Mr．Robinson＇s study．Show your work．

Use a pencil or a pen with black or dark blue ink．Do NOT write your name．Do NOT write outside the box．

Question 6

Continue your response to QUESTION 6 on this page．

\begin{aligned} & \text { Using } E_{i}=\frac{R_{i} g}{n}, \text { expected value has been calcenlated: } \\ & X^{2}=\frac{(110-89.113)^{2}}{110+89.113}+\frac{(36-58.886)^{2}}{36+58.886}+\frac{(69-91.886)^{2}}{69+91.886}+\frac{(85-62.113)^{2}}{85+62.113}=14.527 \end{aligned}

Use a pencil or a pen with black or dark blue ink．Do NOT write your name．Do NOT write outside the box．

Question 6

Continue your response to QUESTION 6 on this page．

(b)

（ii）Assuming the null hypothesis is true，McNemar＇s test statistic， $\chi^{2}$ ，is distributed approximately as a chi－square

statistic with 1 degree of freedom．Determine the p－value for Mr．Robinson＇s test using the test statistic found

in part（c－i）．

p－value for Mr．Robinson＇s test using the test statistic found

in part（c－i）．

p\left(x^{2}>14.527\right)=1.3817 \times 10^{-4}

（d）（i）Using your results in part（c），what should Mr．Robinson conclude for his hypothesis test at a significance

level of 0．05？

(c)

（d）（i）Using your results in part（c），what should Mr．Robinson conclude for his hypothesis test at a significance

level of 0．05？

(d)

（ii）Recall that Mr．Robinson＇s goal was to increase support for the tablet program．Using only the 105 responses

of the parents／guardians whose opinions changed，explain how Mr．Robinson might conclude the pamphlet

[Maximum number: 1]

A chi-square test of homogeneity was conducted to determine whether different racial groups have different distributions in the size of their social networks. The chi-square test statistic and the p-value of the test were 17.43 and 0.037 , respectively. Which of the following is the correct interpretation of the p-value?

Assuming that each racial group has the same distribution in size of social network, the probability of finding a test statistic of 17.43 or smaller is 0.037.

Assuming that each racial group has the same distribution in size of social network, the probability of finding a test statistic of 17.43 or larger is 0.037.

There is a 0.037 probability that the different racial groups have the same distribution of social networks.

There is a 0.037 probability that the different racial groups have different distributions of social networks.

Covered syllabus points

3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence question 1

3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence question 2

Question (a)

Question (b)

Question (c)

Question (d)

3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence question 3