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 25, Problem 18E

(a)

To determine

To find out what are the null and alternative hypotheses.

(a)

Expert Solution
Check Mark

Answer to Problem 18E

  H0:μ1=μ2Ha:μ1μ2

Explanation of Solution

The table of the confined and unconfined groups is given in the question. As we know that the null hypothesis states that the population mean in both groups are the same then the null hypothesis is defined as:

  H0:μ1=μ2

And also the alternative hypothesis states the opposite of the null hypothesis. So, the alternative hypothesis is defined as:

  Ha:μ1μ2

Where we have,

  μ1 : mean alpha wave frequency of the population of unconfined individuals.

  μ2 : means alpha wave frequency of the population of confined individuals.

(b)

To determine

To explain are the assumptions necessary for inference met.

(b)

Expert Solution
Check Mark

Answer to Problem 18E

Yes, the assumptions are necessary for inference met.

Explanation of Solution

Thus, the conditions for the inferences are as follows:

Random condition: It is satisfied because the subjects were randomly assigned to a group.

Independent condition: It is satisfied as we are assuming the inmates are unrelated.

Normal condition: It is not satisfied because one distribution contains an outlier.

Thus, we can say that from above the conditions for random and independent are met but the normal condition is not met. And therefore we can say that the assumptions are necessary for inference met.

(c)

To determine

To perform the appropriate test indicating the formula you used, the calculated value of test statistics, the degree of freedom and the P-value.

(c)

Expert Solution
Check Mark

Answer to Problem 18E

There is sufficient evidence to support the claim of change.

Explanation of Solution

The table of the confined and unconfined groups is given in the question. Thus, the mean of the confined and the unconfined groups can be calculated as follows:

  x¯1=10.7+10.7+10.4+10.9+....+11.2+10.410=10.58x¯2=9.6+10.4+9.7+10.3+....+9+10.910=9.78

And the standard deviation can be calculated as:

  s1=(10.710.58)2+(10.710.58)2+(10.410.58)2+....+(10.410.58)2101=0.4590s2=(9.69.78)2+(10.49.78)2+(9.79.78)2+....+(10.99.78)2101=0.5978

Thus, let us define the null hypotheses as:

  H0:μ1=μ2Ha:μ1μ2

Now, the value of the test statistic is calculated as:

  t=(x¯1x¯2)s12n1+s22n2=10.589.780.4590210+0.5978210=3.357

Thus, the degrees of freedom will be as:

  df=(s12n1+s22n2)2s12n1n11+s22n2n21=16

Thus, the P-value is the number in the column title of table T in appendix F containing the t -value in the row as:

  P<0.01

As we know that the P-value is less than or equal to the significance level then the null hypothesis is rejected. Thus, we have,

  P<0.05Reject H0

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

(d)

To determine

To state your conclusion.

(d)

Expert Solution
Check Mark

Explanation of Solution

The table of the confined and unconfined groups is given in the question. Thus, the P-value is the number in the column title of table T in appendix F containing the t -value in the row as:

  P<0.01

As we know that the P-value is less than or equal to the significance level then the null hypothesis is rejected. Thus, we have,

  P<0.05Reject H0

Thus, we conclude that there is sufficient evidence to support the claim of change. Thus, from part (c) we concluded that there is sufficient evidence to support the claim of change.

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