Question

Grades on a standardized test are known to have a mean of 1020 for students in the United States. The test is administered to 473 randomly selected students in Florida; in this sample, the mean is 1033.26 and the standard deviation (s) is 110.16. The 95% confidence interval for the average test score for Florida students is ( 1023.33, 1043.19 ). (Round your responses to two decimal places.) Is there statistically significant evidence that Florida students perform differently than other students in the United States? The 95% confidence interval for the average test score for Florida students does not include u = 1020, so the null hypothesis that u = 1020 (that Florida students have the same average performance as other students in the United States) can be rejected at the 5% level. Another 523 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 1039.38 with a standard deviation of 96.90. The 95% confidence interval for the change in average test score associated with the prep course is ( -6.82, 19.06 ). (Round your responses to two decimal places.) Is there statistically significant evidence that the prep course helped? O A. No, because the 95% confidence interval for the change in average test score associated with the prep course includes prep - non-prep = 0. OB. No, because the 95% confidence interval for the change in average test score associated with the prep course does not include prep - non-prep = 19.06. OC. Yes, because the 95% confidence interval for the change in average test score associated with the prep course includes prep - Anon-prep = 19.06. OD. Yes, because the 95% confidence interval for the change in average test score associated with the prep course does not include prep - non-prep = 0.

Answer 1

-Sols - Population mean is an average of all the values of population. The confidence interval is a set of values for the population mean. Population mean = M = 1020 Sample Size = n =473 Sample mean= = 1033.26 Sample Standard error = SE ()- 110.16 1473 95% confidence interval foo the average test score es given below: 11.965E (2) Therefore, 110.16 95% contidence integral = 1033.26 311967 1473 110.16 = 1033.26 £1.96. 20 = 21.7486 -1033.26 £1.96€ 5.0652) - 1033.26 49.927792 11023.332208,1043.187792) = (1093.33, 1043.19) The for the average test score 95% Contidence interval (1023.33,1043.19) oss

The 95%. Contidence interval does not include the - average test score of se=1020 so, the null hypothesis HoiM=1000 can't be accepted. (ue. Ho rejected) at 5% level. There is a statistical evidence that the sampled - students of flosida do pestoom didderently to the US Students, Anothed sample of Students are selected at random from florida. Sample size = 523 Sample mean = 1039.38. Parodom Sample Std.emo: SET = SEU 96.90 = 55523 formula for 95% Contidence interval for the difference in average score is given below. [(y-7) 11.96 SECU->)] ?? (96.90)? V 473 523 the Therefore, 95% confidence Pnterval for the in Students average test score is given below bodence interval = (1039.38- 1033.26) +1.96 ( 110. 16) 16.90) 70473 523 where, SE (72) = terval for the difference intervall 95% contidence
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6.12 + 1.96 12135.2256 +9389.61 473 523 6. 12 + 1.96 25.6559 +17.9534 = 6.12 11.96 143.6093 = 6.12 1 1096 (6.6037) 6.12 112.943252 = (-6.823252, 19.063252) & (-6.82, 19.06) The 95%, contidence interval for the difference in students average test score is, - (-6-82, 19.06) The 95% confidence interval does include the average test score ditterence of o, so the null heppothesis, Hool=o con't be sejected. Answeg option (C).. because the 95% confidence interval for the chance test score test score in average associated associated with the prep course includs reprep - non-prep = 19.06. be