Question

The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's in-state and out-of-state applicants. A random sample of 8 in-state applicants results in a SAT scoring mean of 1144 with a standard deviation of 25. A random sample of 17 out-of-state applicants results in a SAT scoring mean of 1200 with a standard deviation of 26. Using this data, find the 90% confidence interval for the true mean difference between the scoring mean for in-state applicants and out-of-state applicants. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval

Answer 1

Explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean score of in-state applicants

x2 = sample mean score of out-of-state applicants

s1 = sample standard deviation for in-state applicants

s2 = sample standard deviation for out-of-state applicants

n1 = number of in-state applicants

n2 = number of out-of-state applicants

For a 90% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (8 - 1) + (17 - 1) = 23

z = 1.714

x1 - x2 = 1144 - 1200 = - 56

Margin of error = z√(s1²/n1 + s2²/n2) = 1.714√(25²/8 + 26²/17) = 18.61

Confidence interval = - 56 ± 18.61