3.16 Grades on a standardized test are known to have a mean of 1000 for stu- den

Question

3.16 Grades on a standardized test are known to have a mean of 1000 for stu- dents in the United States. The test is administered to 453 randomly selected students in Florida; in this sample, the mean is 1013 and the stan- dard deviation (s) is 108. a. Construct a 95% confidence interval for the average test score for Florida students. b. Is there statistically significant evidence that Florida students perform differently than other students in the United States? c. Another 503 students are selected at random from Florida.They are given a 3-hour preparation course before the test is administered. Their average test score is 1019 with a standard deviation of 95. i. Construct a 95% confidence interval for the change in average test score associated with the prep course. ii. Is there statistically significant evidence that the prep course helped? d. The original 453 students are given the prep course and then are asked to take the test a second time. The average change in their te scores is 9 points, and the standard deviation of the change is 60 points. i. Construct a 95% confidence interval for the change in average test scores.

Explanation / Answer

Solution:

Part a

We are given

Xbar = 1013

SD = 108

Sample size n = 453

Confidence level = 95%

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Degrees of freedom = n – 1 = 452

Critical value t = 1.9652

Confidence interval = 1013 -/+ 1.9652*108/sqrt(453)

Confidence interval = 1013 -/+ 9.9721

Lower limit = 1013 – 9.9721 = 1003.03

Upper limit = 1013 + 9.9721 = 1022.97

Confidence interval = (1003.03, 1022.97)

Part b

Yes, there is statistically significant evidence that Florida students perform differently than other students in the United States because the national average of 1000 does not lies between the above confidence interval (1003.03, 1022.97).

Part c.i.

We are given

Xbar = 1019

SD = 95

Sample size = 503

Confidence level = 95%

Degrees of freedom = n – 1 = 503 – 1 = 502

Critical value t = 1.9647

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 1019 -/+ 1.9647*95/sqrt(503)

Confidence interval = 1019 -/+ 8.3222

Lower limit = 1019 – 8.3222 = 1010.68

Upper limit = 1019 + 8.3222 = 1027.32

Confidence interval = (1010.68, 1027.32)

Part c.ii.

Yes, there is statistically significant evidence that the preparation course helped because the confidence interval starts from 1010.68 which is greater than national average of 1000. We reject the null hypothesis that the average is 1000 and we conclude that the average is greater than 1000.

Part d.i.

We are given

Sample size = n = 453

Xbar = 9

SD = 60

Confidence level = 95%

Degrees of freedom = 452

Critical t value = 1.9652

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 9 -/+ 1.9652*60/sqrt(453)

Confidence interval = 9 -/+ 5.5401

Lower limit = 9 – 5.5401 = 3.46

Upper limit = 9 + 5.5401 = 14.54

Confidence interval = (3.46, 14.54)

Part d.ii.

Yes, there is statistically significant evidence that students will perform better on their second attempt after taking the prep course because the difference zero is not lies between above interval.

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