a.
Test whether there is a significant difference in mean time from entry to first stroke for the two entry.
a.
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
2.
The null hypothesis is given below:
Null hypothesis:
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:
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
From Chapter 16 Appendix Table 2, the critical-value for
Rejection rule:
If
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,
Thus, the test statistic is –6.
7.
Conclusion:
Here, the signed-rank sum is greater than the critical value.
That is,
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.
Test whether the data suggest a difference in mean initial velocity for the two entry methods.
b.
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
2.
The null hypothesis is given below:
Null hypothesis:
That is, the mean initial velocity is same for the two entry methods.
3.
The alternative hypothesis is given below:
Alternative hypothesis:
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
From Chapter 16 Appendix Table 2, the critical-value for
Rejection rule:
If
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,
Thus, the test statistic is –33.
7.
Conclusion:
Here, the signed-rank sum is greater than the critical value.
That is,
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.
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Chapter 16 Solutions
INTRODUCTION TO STATISTICS & DATA ANALYS
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