Question

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 American students had a mean height of 67.7 inches with a standard deviation of 3.06 inches. A random sample of 17 non-American students had a mean height of 64.7 inches with a standard deviation of 1.97inches. Determine the 90% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed.

Step 3 of 3 :  

Construct the 90%90% confidence interval. Round your answers to two decimal places.

Answer 1

Given that,

For American : n1 = 12, x1-bar = 67.7 inches and s1 = 3.06 inches

For Non-American:n1=17, x2-bar=64.7 inches and s2=1.97 inches

Since, we assume that the population variances are equal we find the pooled variance first,

Pooled variance is,

Degrees of freedom = 12 + 17 - 2 = 27

Using t-table we get, t-critical value at significance level of 0.10 with 27 degrees of freedom is,

The 90% confidence interval for (μ1 - μ2) is,

Therefore, required 90% confidence interval is (1.41, 4.59)