Last fall a sample of n=26 freshman was selected to participate in a new trainin
ID: 2926001 • Letter: L
Question
Last fall a sample of n=26 freshman was selected to participate in a new training program designed to improve test-taking skills. to evaluate the effectiveness of the new program, the sample was compared with the rest of the freshman class. all freshmen must take the same math class, and the mean score on the final exam for the entire freshman class was u =74. The students in the new program had a mean score of M=78 with SS=2400.
a. On the basis of these data, can the college conclude that the students in the new program performed significantly better than the rest of the freshman class? Use a one tailed test with = .05 Be sure to show all formulas with symbols (and plug in numbers), steps, processes and calculations for all parts of the answers (including how you arrived at the critical value).
b. Can the college conclude that the students in the new program are significantly different from the rest of the freshman class? Use a two-tailed test with = .05 Be sure to show all formulas with symbols (and plug in numbers), steps, processes and calculations for all parts of the answers (including how you arrived at the critical value).
Explanation / Answer
(A)
Below are the null and alternate hypothesis
H0: mu <= 74
H1: mu > 74
variance = SS/(n-1) = 2400/25 = 96
std. dev. = sqrt(96) = 9.8
test statisctics, t = (xbar - mu)/(std.dev./sqrt(n))
t = (78 - 74)/(9.8/sqrt(26))
t = 2.081
p-value = 0.0239 i.e. P(X>78)
As p-value is less than significance level of 0.05, we reject null hypothesis.
Critical value = 1.7081
(B)
Below are the null and alternate hypothesis
H0: mu = 74
H1: mu not equals to 74
variance = SS/(n-1) = 2400/25 = 96
std. dev. = sqrt(96) = 9.8
test statisctics, t = (xbar - mu)/(std.dev./sqrt(n))
t = (78 - 74)/(9.8/sqrt(26))
t = 2.081
p-value = 2*0.0239 = 0.0478
As p-value is less than significance level of 0.05, we reject null hypothesis.
Critical value = 2.059
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