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 18 men had a mean height of 71.4 inches with a standard deviation of 1.68 inches. A random sample of 10 women had a mean height of 65 inches with a standard deviation of 3.01 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

(A) Population variance is unknown, but we are assuming that population variances are equal. so, we will use t distribution.

degree of freedom = n1 +n2 -2

where n1 = 18 and n2= 10

so, degree of freedom = 18+10 -2 = 26

Using t distribution table for degree of freedom 26, we get

t critical value = 2.479 for 98% confidence interval

(B) Population variances are assumed to be equal. So, we will first calculated pooled variance then standard error.

formula for pooled variance =

where n1 = 18, s1 = 1.68, n2 = 10 and s2 =3.01

setting the values, we get

Pooled variance

Formula for standard error is given as

setting the values, we get

So, standard error is 0.88

(C) Formula for confidence interval is given as

x1(bar) = 71.4 and x2(bar) = 65

setting the values, we get

Therefore, required 98% confidence interval is (4.22,8.58)