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 16 in-state applicants results in a SAT scoring mean of 1144 with a standard deviation of 55. A random sample of 10 out-of-state applicants results in a SAT scoring mean of 1185 with a standard deviation of 41. Using this data, find the 99% 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

T-test for two Means Confidence Interval -Unequal Variance

The following information about the sample has been provided: a. Sample Means : Xˉ1​=1144 and Xˉ2​=1185 b. Sample Standard deviation: s1=55 and s2=41 c. Sample size: n1=16 and n2=10 d. Level of confidence is 99% Point estimate = 1144-1185 = -41 The degrees of freedom Assuming that the population variances are unequal, the degrees of freedom are given by Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is 23.0983. Therefore the critical value is tc​=2.8073. This can be found by either using excel or the t distribution table. Standard Error The Standard error is computed as: Confidence Interval The 99% Confidence interval is computed as: Therefore, we are 99% confident that the true difference between population means is contained by the interval (-94.0551, 12.0551)