An English teacher has been teaching a sixth grade composition class for many years. He has the feeling that over the past several years, the writing ability of students has changed. A national test of proficiency in composition was administered 5 years ago. The resulting distribution of scores was normally shaped, had a mean of 85 and a standard deviation of 10.9. In order to test his feeling, he gives his present class of 43 students the same proficiency test. The resulting mean is 80 and the standard deviation is 8.7. Using the z test with a = 0.05 2 tail , what is your conclusion?
A) reject H 0 ; the present class appears deficient in composition writing
B) retain H 0 ; the present class is not deficient in composition writing
C) accept H 0 ; the present class is not deficient in composition writing
D) reject H 0 ; the present class is not deficient in composition writing

Question

Quizplus · Accepted Answer

We need to use the z-test with a 0.05 significance level to compare the mean score of the present class with the mean score of the national test.

z-score = (80 - 85) / (10.9 / sqrt(43)) = -2.30

Looking at the z-table, we find that the area under the normal distribution curve beyond -1.96 and 1.96 is 0.05 (for a two-tail test).

Since the calculated z-score falls beyond -1.96, we reject the null hypothesis (H0: the present class is not deficient in composition writing) and conclude that the present class appears deficient in composition writing.

An English Teacher Has Been Teaching a Sixth Grade Composition