Question

A city council wants to know whether residents of one neighborhood tend to use city parks more often than residents of another neighborhood. After sampling both populations, they find that of 24 residents from neighborhood A, the average number of trips to a park per year was 14, with a standard deviation of 4. For 18 residents in neighborhood B, the average number of trips was 18, with a standard deviation of 8.

Find a 99% confidence interval for the difference in park trips per year, assuming that the sample variances are not necessarily equal.

Find a 95% confidence interval assuming that the variances are equal.

Answer 1

To find the 99% confidence interval, use a two-sample t-test. For the 95% confidence interval assuming equal variances, use the pooled standard deviation.

Step-by-step explanation:

To find a 99% confidence interval for the difference in park trips per year between the two neighborhoods, we can use the two-sample t-test since the sample sizes are small. First, we calculate the standard error of the difference in means, which is the square root of the sum of the variances divided by the sample sizes. For the 99% confidence interval, we use the t-distribution with degrees of freedom equal to the smaller of n1-1 and n2-1. The critical value for a one-tail test at a 99% confidence level with 40 degrees of freedom is approximately 2.704.

For the 95% confidence interval assuming equal variances, we can use the pooled standard deviation to calculate the standard error of the difference in means. The formula for the pooled standard deviation is calculated by taking the weighted average of the sample variances, where the weights are the degrees of freedom for each sample. The critical value for a two-tail test at a 95% confidence level with 41 degrees of freedom is approximately 2.021.