Book edition OER 2018

Author(s) Barbara Illowsky, Susan Dean

Pages 902 pages

ISBN 9781938168208

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Short Answer

Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics and Asians? Suppose that Table 13.26 shows the results of a study.
White Black Hispanics Asian
6 4 7 8
8 1 3 3
2 5 5 5
4 2 4 1
6 6 7
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of $0.05$ .

We accept the null hypothesis.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

White	Black	Hispanics	Asian
6	4	7	8
8	1	3	3
2	5	5	5
4	2	4	1
6		6	7

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of $0.05$ .

Step 2: Explanation

Consider the data provided in the table13.26, to test mean numbers of times a person eat outs are same at $5 %$ level of significance:

Consider the null hypothesis that there is no significance difference between the mean numbers of times a person eat outs. For this, use Ti-83 calculator. For this, click on STAT $\to$ press $1 \to$ EDIT, then put the above given weights of different groups data into the list $L 1, L 2, L 3 a n d L 4$ . The screenshot is given as below:

Step 3: Explanation

The calculator displays :

$F = 0.6388149$

$p = 0.5437096$

Factor

$d f = 2$

$S S = 20.17321$

$M S = 10.0866$

Error

$d f = 13$

S S = 205.264286

$M S = 15.7895604$

Step 4: Explanation

We use a solution sheet to conduct the hypothesis tests, and we have:

a) The null hypothesis that three mean commuting mileages are the same is:

$H_{0} : μ_{o} = μ_{h} = μ_{f}$

b) The alternate hypothesis is that at least any two of the means are different.

c) The degree of freedom in the numerator - $d f (n u m)$ is localid="1649157482240" $2$ ,

and the degree of freedom in the denominator - $d f (d e n o m)$ is $13$ .

d) We use the $F$ distribution for the test.

e) The value of the test statistic $(F - v a l u e)$ is $0.6388$ .

f) The $P$ -value for the test is $0.5437$ .

Step 5: Explanation

g). The graph of the distribution is

i. Level of significance $α$ is $0.05$ .

ii. Decision: We do not reject the null hypothesis

iii. Reason for decision: $P$ -value is $0.5437$ which is greater than the $0.05$ level of significance.

iv. Conclusion: The mean scores of different class delivery are not different.

Step 6: Final Answer

We accept the null hypothesis.

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