Ackerman and Goldsmith (2011) found that students who studied text from printed hard copy had better test scores than students who studied from text presented on a screen. In a related study, a professor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was μ = 81.7, but the n = 9 students who used e-books had a mean of M = 77.2 with a standard deviation of s = 5.7. Use the Distributions tool to answer the questions that follow. t Distribution Degrees of Freedom = 21
t Distribution
Degrees of Freedom = 21
The null and alternative hypotheses are given below:
Null Hypothesis:
H0: Scores for students with e-books are not significantly different.
Alternative Hypothesis:
H1: Scores for students with e-books are significantly different.
Given information:
Test statistic:
p value:
degrees of freedom=n-1=9-1=8
test statistic=-2.3684
p value=0.0454, obtained from the excel function, =T.DIST.2T(2.3684,8).
Decision Rule:
If p-value ≤ α, then reject the null hypothesis.
Conclusion:
Let the level of significance is α=0.05
Here, the p-value is lesser than the level of significance.
From the decision rule, reject the null hypothesis.
It can be concluded that “Scores for students with e-books are significantly different”.
Correct option: Reject the null hypothesis; scores for students with e-books are significantly different.
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