Exhibit 10.9.A tutor promises to improve GMAT scores of students by more than 50 points after three lessons.To see if this is true,the tutor takes a sample of 49 students' test scores after and before they received tutoring.The mean difference was 53 points better after tutoring,with a standard deviation of the difference equal to 12 points.Let denote the mean of the difference: score after tutoring minus score before tutoring. Refer to Exhibit 10.9.What is the appropriate conclusion at the 5% significance level?
A)Reject ,we can conclude that the mean difference in test scores after and before tutoring is greater than 50 points.
B)Reject ,we cannot conclude that the mean difference in test scores after and before tutoring is greater than 50 points.
C)Fail to reject ,we can conclude that the mean difference in test scores after and before tutoring is greater than 50 points.
D)Fail to reject

Question

Exhibit 10.9.A tutor promises to improve GMAT scores of students by more than 50 points after three lessons.To see if this is true,the tutor takes a sample of 49 students' test scores after and before they received tutoring.The mean difference was 53 points better after tutoring,with a standard deviation of the difference equal to 12 points.Let  denote the mean of the difference: score after tutoring minus score before tutoring. Refer to Exhibit 10.9.What is the appropriate conclusion at the 5% significance level?
A)Reject ,we can conclude that the mean difference in test scores after and before tutoring is greater than 50 points. 
B)Reject ,we cannot conclude that the mean difference in test scores after and before tutoring is greater than 50 points. 
C)Fail to reject ,we can conclude that the mean difference in test scores after and before tutoring is greater than 50 points. 
D)Fail to reject

Quizplus · Accepted Answer

We need to test the null hypothesis that the mean difference in test scores after and before tutoring is not greater than 50 points. 
Ho: μ ≤ 50 
Ha: μ > 50 
We will use a one-tailed t-test with a significance level of 0.05, since the tutor promised to improve scores by more than 50 points. 
t = (53 - 50) / (12 / sqrt(49)) = 3.06 
The critical value for a one-tailed test with 48 degrees of freedom is 1.677. Since 3.06 > 1.677, we reject the null hypothesis. Therefore, we can conclude that the mean difference in test scores after and before tutoring is greater than 50 points.