Question

) Grades on a standardized test are known to have a mean of 1000 for students in the US. The test is administered to 453 randomly selected students from Queens College, and they obtained an average grade of 1013 and a standard deviation of 108. a. Construct a 95 % confidence interval for the true average test score for Queens College students.(1pt) b. With 5% significance level, Is there a statistical evidence that Queens College students perform differently than other students in the US?(1pt) From now, assume the same 453 students selected earlier are now given a two-hour tutoring session and then asked to take the test a second time. The average change in their test scores is 9 points, and the standard deviation of the change is 60 points. Assume the changes are from a Normal ( , ) 2   distribution, and for every student, the change in score is the difference between the score after the tutoring session, and the score before the tutoring session c. You are asked by the school administration to study whether students have performed better in their second attempt. State in terms of  , the relevant null and alternative hypothesis in conducting this study.d. Compute the t statistic for testing ?0 against ??; obtain the p- value for the test.e. Do you reject ?0 at the 5% level? At the 1% level? f. Provide a short summary of your conclusions from this study. Comment on the practical versus statistical significance of this estimate.

Answer 1

a.

Standard error of mean = = = 5.074283

Since the sample size is large, we can assume the critical t statistic for 95% confidence interval to be 1.96

95 % confidence interval for the true average test score for Queens College students is,

(1013 - 1.96 * 5.074283 ,  1013 + 1.96 * 5.074283)

(1003.054 , 1022.946)

(b)

Since the 95 % confidence interval for the true average test score for Queens College students does not contain the value 1000, there is a statistical evidence that Queens College students perform differently than other students in the US at 5% significance level.

(c)

Null Hypothesis H0: The difference between the mean scores after the tutoring session, and the score before the tutoring session is zero.

Alternative Hypothesis Ha: The difference between the mean scores after the tutoring session, and the score before the tutoring session is greater than zero.

(d)

Standard error of mean difference = = = 2.819046

Test statistic, t = mean difference / Standard error of mean difference = 9 / 2.819046 = 3.19

(e)

Degree of freedom = n-1 = 453-1 = 452

P-value = P(t > 3.19) = 0.00076

Since, p-value is less than 0.05 significance level, we reject null hypothesis H0 at 5% significance level.

Since, p-value is less than 0.01 significance level, we reject null hypothesis H0 at 1% significance level.

(f)

Since we reject null hypothesis H0 , we conclude that there is strong evidence that the difference between the mean scores after the tutoring session, and the score before the tutoring session is greater than zero.

The mean difference of 9 points with population mean of 1000 does not semms to have practical significance.