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Name: Ackerman and Goldsmith (2011) found that students who studied text fro
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Description: 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

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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

-3.0-2.0-1.00.01.02.03.0x

Is the sample sufficient to conclude that scores for students using e-books were significantly different from scores for the regular class? Use a two-tailed test with α = .05.

Reject the null hypothesis; scores for students with e-books are significantly different.

Reject the null hypothesis; scores for students with e-books are not significantly different.

Accept the null hypothesis; scores for students with e-books are significantly different.

Accept the null hypothesis; scores for students with e-books are not significantly different.

Construct the 90% confidence interval to estimate the mean exam score if the entire population used e-books:

73.666 to 80.734

75.300 to 79.100

75.340 to 79.060

Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report.

The results show that exam scores significantly different for students using e–books than for other students,

Expert Solution

Step 1

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:

$μ = 81.7 n = 9 M = 77.2 s = 5.7$

Test statistic:

$t = \frac{M - μ}{\frac{s}{\sqrt{n}}} = \frac{77.2 - 81.7}{\frac{5.7}{\sqrt{9}}} = - 2.3684$

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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