Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
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Chapter 24, Problem 31E
To determine

To explain is there any evidence that the mean time to finish is different for randomized heats.

Expert Solution & Answer
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Answer to Problem 31E

No, there is not any evidence that the mean time to finish is different for randomized heats.

Explanation of Solution

Let us the check the conditions and assumptions for inference for this case as:

Random condition: It is satisfied because the heats were randomized.

Independent condition: It is satisfied because since no individual can be in more than on heat.

Normal condition: It is not satisfied because the boxplot contains an outlier and the other distribution is strongly skewed.

Thus, all the conditions are not met.

Now, it is given that:

  x¯1=52.3557x¯2=52.3286n1=7n2=7s1=1.6932s2=1.2005

The mean and the standard deviation is calculated by using excel by going to the data tab and then to the analysis tab and then selecting the data analysis option. Then we go to the t -test option in the dialogue box that appeared on screen. Then we select the data and the result will come out.

Now, let us define the hypotheses for testing:

  H0:μ1=μ2H1:μ1μ2

Thus, the value of test statistics is:

  t=x¯1x¯2s12n1+s22n2=52.355752.32861.693227+1.200527=0.03

And the value of degree of freedom is as:

  df=(s12n1+s22n2)(s12n1)2n11+(s22n2)2n21=10

Thus, the P-value will be as:

  P>0.20

And as we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected. So,

  P>0.05Fail to Reject H0

Thus, we conclude that there is not sufficient evidence to support the claim of a difference.

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