Chapter 16.2, Problem 14E

a.

To determine

Test whether there is a significant difference in mean time from entry to first stroke for the two entry.

Answer to Problem 14E

The conclusion is that there is no significant difference in mean time from entry to first stroke for the two entry.

Explanation of Solution

The data based on the water entry time to the first stroke for Flat and hole.

1.

Consider ${\mu }_d$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H_0:{\mu }_d=0$

That is, the mean time from entry to first stroke is same for the two entry methods.

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$H_a:{\mu }_d\ne 0$

That is, the mean time from entry to first stroke is not same for the two entry methods.

4.

Test statistic:

Here, the test statistic is signed-rank sum.

5.

Critical value:

Here, the test is one-tailed test with $n=10$.

From Chapter 16 Appendix Table 2, the critical-value for $n=10$ with $\alpha =0.01$ is 45.

Rejection rule:

If $\text{Signed-rank sum}\ge 45\left(=\text{Critical value}\right)$ or $\text{Signed-rank sum}\le -45\left(=-\text{Critical value}\right)$, then the null hypothesis is rejected.

6.

Calculation:

The difference is obtained below:

Swimmer	Hole	Flat	Difference
1	1.18	1.06	0.12
2	1.1	1.23	–0.13
3	1.31	1.2	0.11
4	1.12	1.19	–0.07
5	1.12	1.29	–0.17
6	1.23	1.09	0.14
7	1.27	1.09	0.18
8	1.08	1.33	–0.25
9	1.26	1.27	–0.01
10	1.27	1.38	–0.11

Ordering the absolute differences results in the following assignment of signed ranks.

Difference	Signed Rank
0.01	–1
0.07	–2
0.11	–3.5
0.11	3.5
0.12	5
0.13	–6
0.14	7
0.17	–8
0.18	9
0.25	–10

The test statistic is,

$\begin{array}{c}\text{Signed-rank sum}=-1-2-3.5+3.5+5-6+7-8+9-10\\ =-6\end{array}$

Thus, the test statistic is –6.

7.

Conclusion:

Here, the signed-rank sum is greater than the critical value.

That is, $-6\left(\text{=Signed-rank sum}\right)\ge -45\left(=-\text{Critical value}\right)$.

By the rejection rule, the null hypothesis is not rejected.

Thus, there is no significant difference in mean time from entry to first stroke for the two entry.

b.

To determine

Test whether the data suggest a difference in mean initial velocity for the two entry methods.

Answer to Problem 14E

The conclusion is that there is no evidence that the data suggest a difference in mean initial velocity for the two entry methods.

Explanation of Solution

The data based on the initial velocity for the Flat and hole.

1.

Consider ${\mu }_d$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H_0:{\mu }_d=0$

That is, the mean initial velocity is same for the two entry methods.

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$H_a:{\mu }_d\ne 0$

That is, the mean initial velocity is not same for the two entry methods.

4.

Test statistic:

Here, the test statistic is signed-rank sum.

5.

Critical value:

Here, the test is one-tailed test with $n=10$.

From Chapter 16 Appendix Table 2, the critical-value for $n=10$ with $\alpha =0.01$ is 45.

Rejection rule:

If $\text{Signed-rank sum}\ge 45\left(=\text{Critical value}\right)$ or $\text{Signed-rank sum}\le -45\left(=-\text{Critical value}\right)$, then the null hypothesis is rejected.

6.

Calculation:

The difference is obtained below:

Swimmer	Hole	Flat	Difference
1	24	25.1	–1.1
2	22.5	22.4	0.1
3	21.6	24	–2.4
4	21.4	22.4	–1
5	20.9	23.9	–3
6	20.8	21.7	–0.9
7	22.4	23.8	–1.4
8	22.9	22.9	0
9	23.3	25	–1.7
10	20.7	19.5	1.2

Ordering the absolute differences results in the following assignment of signed ranks.

Difference	Signed Rank
0	-
0.1	1
0.9	–2
1	–3
1.1	–4
1.2	5
1.4	–6
1.7	–7
2.4	–8
3	–9

The test statistic is,

$\begin{array}{c}\text{Signed-rank sum}=1-2-3-4+5-6-7-8-9\\ =-33\end{array}$

Thus, the test statistic is –33.

7.

Conclusion:

Here, the signed-rank sum is greater than the critical value.

That is, $-33\left(\text{=Signed-rank sum}\right)\ge -45\left(=-\text{Critical value}\right)$.

By the rejection rule, the null hypothesis is not rejected.

Thus, there is no evidence that the data suggest a difference in mean initial velocity for the two entry methods.