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 20E

(a)

To determine

To test appropriate hypotheses and state your conclusion.

(a)

Expert Solution
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Explanation of Solution

In the question, the mortality rate and the derby of north and south data is collected for both and the box plot for the same is given. It is also given that:

  x¯1=1631.59x¯2=1388.85n1=34n2=27s1=138.470s2=151.114

Now, let us define the hypotheses for testing:

  H0:μ1=μ2H1:μ1μ2

Thus, the value of test statistic is:

  t=x¯1x¯2s12n1+s22n2=1631.591388.85138.470234+151.114227=6.47

And the degrees of freedom is:

  df=(s12n1+s22n2)(s12n1)2n11+(s22n2)2n21=53

Therefore, the P-value will be:

  P<0.01

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<0.05Reject H0

Thus, we conclude that there is a sufficient evidence to support the claim of a difference.

(b)

To determine

To explain what effect might the outlier have had on your test.

(b)

Expert Solution
Check Mark

Explanation of Solution

In the question, the mortality rate and the derby of north and south data is collected for both and the box plot for the same is given. And there is a sufficient evidence to support the claim of a difference. So, the outlier might have had effect on our test that the large outlier would have increased both the mean and the standard deviation of the North group, which could then exaggerate the actual differences between the two groups and thus it is possible that there actually is no significant difference (if the outlier is not taken into account).

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