AP Statistics 8.6 Carrying Out a Chi-Square Test for Homogeneity or Independence- MCQs - Exam Style Questions
Question
A college offers both two-year and four-year programs of study. An administrator is investigating whether there is an association between the program of study and the length of the program. A random sample of students in all programs offered by the college was selected. All randomly selected students indicated their program of study and the length of the program. The results obtained are shown in the table.
Program of Study by Length of the Program
Program of Study | Two-Year | Four-Year | Total |
---|---|---|---|
Liberal arts | 123 | 87 | 210 |
Business | 315 | 145 | 460 |
Health | 156 | 65 | 221 |
Social sciences | 214 | 143 | 357 |
Total | 808 | 440 | 1,248 |
The administrator wants to test the following hypotheses:
\(H_0:\) There is no association between the program of study and length of the program.
\(H_a:\) There is an association between the program of study and length of the program.
Assuming the conditions for inference have been met, the resulting test statistic was \( \chi^2 \approx 13.23 \).
\(H_0:\) There is no association between the program of study and length of the program.
\(H_a:\) There is an association between the program of study and length of the program.
Assuming the conditions for inference have been met, the resulting test statistic was \( \chi^2 \approx 13.23 \).
At the 1 percent level of significance, which of the following is an appropriate conclusion of the test?
(A) The null hypothesis should not be rejected, and there is convincing statistical evidence of no association between the program of study and length of the program.
(B) The null hypothesis should not be rejected, and there is convincing statistical evidence of an association between the program of study and length of the program.
(C) The null hypothesis should not be rejected, and there is not convincing statistical evidence of an association between the program of study and length of the program.
(D) The null hypothesis should be rejected, and there is convincing statistical evidence of no association between the program of study and length of the program.
(E) The null hypothesis should be rejected, and there is convincing statistical evidence of an association between the program of study and length of the program.
(B) The null hypothesis should not be rejected, and there is convincing statistical evidence of an association between the program of study and length of the program.
(C) The null hypothesis should not be rejected, and there is not convincing statistical evidence of an association between the program of study and length of the program.
(D) The null hypothesis should be rejected, and there is convincing statistical evidence of no association between the program of study and length of the program.
(E) The null hypothesis should be rejected, and there is convincing statistical evidence of an association between the program of study and length of the program.
▶️ Answer/Explanation
Detailed solution
Degrees of freedom: \((r-1)(c-1)=(4-1)(2-1)=3\).Critical value at \(\alpha=0.01\) and \(df=3\): \(\chi^2_{0.01,3}\approx 11.34\).
Since \(13.23 > 11.34\), the p-value \(<0.01\) ⇒ reject \(H_0\).
Conclusion: there is convincing statistical evidence of an association between program of study and length of program.
✅ Answer: (E)
Question
A polling agency conducted a survey about social media in which each person in random samples of \(1{,}000\) men and \(1{,}000\) women was asked what factor he or she considers to be the most important when deciding whether to connect on social media with another person. The responses are shown in the table.
Personal Friend | Stay in Touch | Mutual Friends | Business Networking | Other | |
---|---|---|---|---|---|
Men | \(600\) | \(210\) | \(105\) | \(45\) | \(40\) |
Women | \(650\) | \(224\) | \(65\) | \(15\) | \(46\) |
What is the contribution to the chi-square test statistic for men who selected business networking as the most important factor?
(A) \(0.5\)
(B) \(5\)
(C) \(7.5\)
(D) \(30\)
(E) \(45\)
(B) \(5\)
(C) \(7.5\)
(D) \(30\)
(E) \(45\)
▶️ Answer/Explanation
Detailed solution
We need the single-cell contribution \(\displaystyle \frac{(O-E)^2}{E}\) for Men × Business Networking.
Row total for men: \(1{,}000\). Column total for business networking: \(45+15=60\). Table total: \(2{,}000\).
Expected count: \(\displaystyle E=\frac{(1{,}000)(60)}{2{,}000}=30\). Observed count: \(O=45\).
Contribution: \(\displaystyle \frac{(45-30)^2}{30}=\frac{225}{30}=7.5\).
✅ Answer: (C)