Question

1. males and females were asked about what they would do if they received a $100 bill by mail, addressed to their neighbor, but wrongly delivered to them. would they return it to their neighbor? of the 69 males randomly sampled, 52 said yes and of the 131 females randomly sampled, 120 yes. a) does the data indicate that the proportions that said yes are different for male and female at a 5% level of significance? b) use a 95% confidence interval to know if there is a difference in the responses of males and females. c) what are the assumptions required for the ci to be valid?

Explanation:

Step1: State the hypotheses

Let \(p_m\) be the proportion of males who would return the bill and \(p_f\) be the proportion of females who would return the bill.
\(H_0:p_m = p_f\) (null hypothesis)
\(H_a:p_m
eq p_f\) (alternative hypothesis)

Step2: Calculate the sample proportions

For males: \(n_m = 69\), \(x_m=52\), \(\hat{p}_m=\frac{52}{69}\approx0.75\)
For females: \(n_f = 131\), \(x_f = 120\), \(\hat{p}_f=\frac{120}{131}\approx0.92\)
The pooled proportion \(\hat{p}=\frac{x_m + x_f}{n_m + n_f}=\frac{52+120}{69 + 131}=\frac{172}{200}=0.86\)

Step3: Calculate the test - statistic

The formula for the test - statistic \(z\) in a two - proportion z - test is \(z=\frac{\hat{p}_m-\hat{p}_f}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_m}+\frac{1}{n_f})}}\)
Substitute the values:

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Step4: Find the p - value

Since this is a two - tailed test, the p - value is \(2P(Z\lt - 3.29)\)
From the standard normal table, \(P(Z\lt - 3.29)=0.0005\), so the p - value \(=2\times0.0005 = 0.001\)

Step5: Make a decision

Given \(\alpha = 0.05\), since \(p - value=0.001\lt0.05\), we reject \(H_0\)

Answer:

There is sufficient evidence at the \(5\%\) level of significance to conclude that the proportions of males and females who would return the bill are different.