Question

A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 10 men had a mean height of 71.3 inches with a standard deviation of 2.34 inches. A random sample of 17 women had a mean height of 65.5 inches with a standard deviation of 2.72 inches. Determine the 98% confidence interval for the true mean difference between the mean height of the men and the mean height of the women. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3:

Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3:

Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places.

Step 3 of 3:

Construct the 98% confidence interval. Round your answers to two decimal places.

Answer 1

we want to calculate 98% confidence interval for the true mean difference between the mean height of the men and the mean height of the women

Given that

and

Confidence level = C= 0.98

Given that population variances are equal and two populations are normally distributed

Therfore 98% confidence interval for the true mean difference is

Where tc = Critical value of C = 2.485 ..............(By using t table)

Degrees of freedom = n1+n2-2 = 10+17-2 = 25

and standard error =

Where Pooled SD =

therore  

we get Pooled SD=

Since we assume that the population variances are equal, the standard error is computed as follows:

we get standard error = Se = 2.032 ....................(Answer)

Therefore the 98% Confidence interval for mean difference is

Therfore 98% confidence interval for the true mean difference is

The 98% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is ............................(Answer)