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

Question

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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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Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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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Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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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Answer 5

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

Answer 6

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.

a.

Expert Solution

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 $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

Answer 7

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 8

a.

Expert Solution

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.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

1.

Consider $μd$ is the mean difference in the time from the water entry to first stroke.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=−1−2−3.5+3.5+5−6+7−8+9−10=−6$

Thus, the test statistic is –6.

7.

Conclusion:

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

That is, $−6(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

Answer 13

b.

To determine

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

b.

Expert Solution

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 $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

Answer 14

b.

To determine

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

Answer 15

b.

Expert Solution

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.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

1.

Consider $μd$ is the difference in mean time from entry to first stroke for the two entry.

2.

The null hypothesis is given below:

Null hypothesis:

$H0:μd=0$

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

3.

The alternative hypothesis is given below:

Alternative hypothesis:

$Ha:μd≠0$

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 $α=0.01$ is 45.

Rejection rule:

If $Signed-rank sum≥45(=Critical value)$ or $Signed-rank sum≤−45(=−Critical value)$ , 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,

$Signed-rank sum=1−2−3−4+5−6−7−8−9=−33$

Thus, the test statistic is –33.

7.

Conclusion:

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

That is, $−33(=Signed-rank sum)≥−45(=−Critical value)$ .

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.

Answer 20

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

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

Answer to Problem 14E

Explanation of Solution

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

Answer to Problem 14E

Explanation of Solution

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