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Expert-verified Found in: Page 771 ### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208 # 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 Asian6478813325554241667Assume 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.

See the step by step solution

## 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 $L1,L2,L3andL4$. The screenshot is given as below: ## Step 3: Explanation

The calculator displays :

$F=0.6388149$

$p=0.5437096$

Factor

$df=2$

$SS=20.17321$

$MS=10.0866$

Error

$df=13$

$SS=205.264286$

$MS=15.7895604$

## 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:

${H}_{0}:{\mu }_{o}={\mu }_{h}={\mu }_{f}$

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

c) The degree of freedom in the numerator - $df\left(num\right)$ is localid="1649157482240" $2$,

and the degree of freedom in the denominator - $df\left(denom\right)$ is $13$.

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

e) The value of the test statistic $\left(F-value\right)$ is $0.6388$.

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

## Step 5: Explanation

g). The graph of the distribution is h)

i. Level of significance $\alpha$ 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.

We accept the null hypothesis. ### Want to see more solutions like these? 