Question

In a statistics course project, students are interested in the proportions of 1st year and 4th year students at UW-Madison students who have remained in Madison (in an apartment or in a residence hall) and the proportion that have returned home after courses moved to an online format after spring break. The following data shows the results of a survey. Analyze the data as if the respondents to the survey represent random samples from the populations of 1st and 4th year UW-Madison students. stayed returned 1st year 19 68 4th year 36 17 15 Find a 95% confidence interval for the difference in population proportions who returned home after spring break using a formula. Interpret the confidence interval in context.

Answer 1

The 95% confidence interval for the difference in population proportions who returned home after spring break is (0.31, 0.61).

Explanation:

The (1 - α)% confidence interval for the difference between two population proportions is:

`CI=(\hat p_(1)-\hat p_(2))\pm z_(\alpha/2)* \sqrt{(\hat p_(1)(1-\hat p_(1)))/(n_(1))+(\hat p_(2)(1-\hat p_(2)))/(n_(2))}}`

The information provided is:

Stayed Returned

1st year 19 68

4th year 36 17

Compute the sample proportion of 1st year students who returned home as follows:

`\hat p_(1)=(68)/(68+19)=0.782`

Compute the sample proportion of 4th year students who returned home as follows:

`\hat p_(2)=(17)/(17+36)=0.321`

The critical value of z for 95% confidence level is:

`z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96`

Compute the 95% confidence interval for the difference between two proportions as follows:

`CI=(\hat p_(1)-\hat p_(2))\pm z_(\alpha/2)* \sqrt{(\hat p_(1)(1-\hat p_(1)))/(n_(1))+(\hat p_(2)(1-\hat p_(2)))/(n_(2))}}`

`=(0.782-0.321)\pm 1.96* \sqrt{(0.782(1-0.782))/(87)+(0.321(1-0.321))/(53)}}`

`=0.461\pm (1.96* 0.078)\\=0.461\pm 0.1529\\=(0.3081, 0.6139)\\\approx (0.31, 0.61)`

Thus, the 95% confidence interval for the difference in population proportions who returned home after spring break is (0.31, 0.61).

The 95% confidence interval for difference between two population proportions, (0.31, 0.61) implies that there is a 0.95 probability that the true difference between the proportions is included in the interval.