Chapter 13.3, Problem 5E

a.

To determine

To state: The test hypotheses.

To identify: The claim for the given situation.

Answer to Problem 5E

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0$: There is no significant difference in the scores obtained by freshmen and transfers.

Alternative hypothesis $H_1$:

$H_1$: The scores obtained by freshmen and transfers differ significantly.

The claim is that “There is a difference in the scores obtained by freshmen and transfers”.

Explanation of Solution

Given info:

The data represents the scores obtained by randomly selected freshmen and transfers in order to measure the technical efficiency. The level of significance is $\alpha =0.05$.

Justification:

Null hypothesis states that there is no significant difference in the scores obtained by freshmen and transfers.

Alternative hypothesis states that the scores obtained by freshmen and transfers differ significantly.

The claim is that “There is a difference in the scores obtained by freshmen and transfers”.

b.

To determine

To find: The critical value.

Answer to Problem 5E

The critical value is $\pm 1.96$.

Explanation of Solution

Calculation:

Critical value:

Here, the test is two tailed test.

For the level of significance $\alpha =0.05$,

From “Appendix Table E, Cumulative standard normal distribution”, the critical value that corresponds to two tailed test and level of significance $\alpha =0.05$ is $\pm 1.96$.

Thus, the critical value is $\pm 1.96$.

c.

To determine

To find: The test statistic value.

Answer to Problem 5E

The test statistic is –0.53.

Explanation of Solution

Calculation:

Test statistic:

The test value of the wilcoxon rank sum test is obtained as follows:

The first step for finding the test value is, to combine the data and arrange the data in ascending order then to rank the data indicating the corresponding group.

The rank of the score indicating the corresponding group is,

Score	Group	Rank
29	F	1
30	F	2
32	F	3.5
32	F	3.5
33	T	5
35	F	6.5
35	T	6.5
36	T	8.5
36	T	8.5
37	T	10
38	F	11.5
38	T	11.5
39	F	13.5
39	F	13.5
40	F	15.5
40	F	15.5
41	T	17
43	T	18
44	T	19
45	T	20
46	T	21
47	F	22

The sum of the ranks of the freshmen group is,

$\begin{array}{c}{R}_1=1+2+3.5+3.5+6.5+11.5+13.5+1.35+15.5+15.5+22\\ =108\end{array}$

The sum of the ranks of the transfers group is,

$\begin{array}{c}{R}_2=5+6.5+8.5+8.5+10+11.5+17+18+19+20+21\\ =145\end{array}$

The sum of the ranks with the smaller sum is considered as R.

That is,

$\begin{array}{c}R=\min \left(R_1,R_2\right)\\ R=\min \left(108,145\right)\\ =108\end{array}$

Thus, the sum of ranks is $R=108$

The mean of the sum of ranks is,

$\begin{array}{c}{\mu }_R=\frac{n_1\left(n_1+n_2+1\right)}{2}\\ =\frac{11\times \left(11+11+1\right)}{2}\\ =\frac{253}{2}\\ =126.5\end{array}$

The standard deviation of sum of ranks is,

$\begin{array}{c}{\sigma }_R=\sqrt{\frac{n_1{n}_2\left(n_1+n_2+1\right)}{12}}\\ =\sqrt{\frac{\left(11\times 11\right)\left(11+11+1\right)}{12}}\\ =\sqrt{\frac{2,783}{12}}\\ =15.2288\end{array}$

The test statistic value is,

$\begin{array}{c}z=\frac{R-{\mu }_R}{{\sigma }_R}\\ =\frac{108-126.5}{15.2288}\\ =\frac{-18.5}{15.2288}\\ =-1.2148\end{array}$

Hence, the test value is $z=-1.2148$

d.

To determine

Whether the null hypothesis is rejected for the given situation.

Answer to Problem 5E

The null hypothesis $H_0$ should not be rejected.

Explanation of Solution

Justification:

De Decision rule for a two tailed test:

• If   $z<-z_0\left(=-1.96\right)$, then reject the null hypothesis
• If $z>z_0\left(=+1.96\right)$, then reject the null hypothesis

Conclusion:

The value of test statistic is -1.2148 and the critical value is -1.96.

Here, the test statistic value is greater than the critical value.

That is, $-1.2148\left(=z\right)>-1.96\left(=-z_0\right)$

Therefore, the decision is “not to reject the null hypotheses”.

Thus, the null hypothesis is not rejected.

e.

To determine

To summarize: The result.

Answer to Problem 5E

There is no evidence to support the claim that the scores obtained by freshmen and transfers differ significantly

Explanation of Solution

From part (d), the null hypothesis is not rejected.

Hence, there is no enough evidence to support the claim that scores obtained by freshmen and transfers differ significantly.

Here, the null hypothesis is not rejected. Thus, there is no evidence to support the claim that there is a difference in the number of credits community college students transferred at 5% level of significance.