To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Fred needs to choose a password for a certain website. Assume that he will choose an 8-character password, and that the legal characters are the lowercase letters a, b, c, ..., z, the uppercase letters A, B, C, ..., Z, and the numbers 0, 1, . . ., 9. (a) How many possibilities are there if he is required to have at least one lowercase letter in his password? (b) How many possibilities are there if he is required to have at least one lowercase letter and at least one uppercase letter in his password? (c) How many possibilities are there if he is required to have at least one lowercase letter, at least one uppercase letter, and at least one number in his password?

a =1500, b=1700 what is percentage of a is b

A 12-inch bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let Y = the distance from the left end at which the break occurs. Suppose Y has the following pdf. f(y) = { (a) Compute the cdf of Y. F(y) = 0 0 y -옴) 0 ≤ y ≤ 12 1- 12 y 12 Graph the cdf of Y. F(y) 1.0 0.8 0.6 0.4 0.2 y 2 6 8 10 12 F(y) F(y) F(y) 1.01 1.0ㅏ 1.0 0.8 0.6 0.4 0.2 0.8 0.8 0.6 0.4 ཨཱུ སྦེ 0.6 0.4 0.2 2 4 6 8 10 12 (b) Compute P(Y ≤ 5), P(Y > 6), and P(5 ≤ y ≤ 6). (Round your answers to three decimal places.) P(Y ≤ 5) = P(Y > 6) = P(5 ≤ y ≤ 6) = (c) Compute E(Y), E(y²), and V(Y). E(Y) = in E(Y2) v(x) = in 2 2 2 4 6 8 10 12 y 2 4 6 8 10 12

Answer 2

Question

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Answer 6

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

Answer 7

Chapter 7.2, Problem 58E

Section 1:

To determine

To test: The significance test to test whether there is a difference in sprint speed of Elite players from the Canadian National team and a university squad.

Answer 8

Section 1:

Expert Solution

Answer to Problem 58E

Solution: There is a significant difference between the sprint speed of both types of players. The t -statistic is 2.89 which lies between P=0.01 and P=0.02 for 12 degrees of freedom.

Answer 9

Section 1:

Expert Solution

Answer 10

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in sprint speed of elite players and a university squad against the alternative that there is difference in the sprint speeds of the elite players and the university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(27.3−26)−00.7216+1.5213=1.30.45=2.89

The P-value for the provided one-sided test is P(T≥2.89). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t-statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=2.89 lies between P=0.005 and P=0.01. As the alternative is two-sided, double the tail area p. Hence, the value t=2.89 lies between P=0.01 and P=0.02.

Answer 13

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Answer 14

To determine

To explain: The conclusion of the performed significance test.

Answer 15

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the sprint speed of the elite players of Canadian National team and a university squad.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The t-test statistic is obtained as 2.89 in the previous part. For 12 degrees of freedom, the Table-D shows that the P lies between 0.01 and 0.02. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggest that there is a significant difference between the sprint speeds of both the types of players.

Answer 20

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

Answer 21

Section 2:

To determine

To test: The Significance test to test whether there is difference in peak heart rate of Elite players from the Canadian National team and the university squad.

Answer 22

Section 2:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad. The t-statistic is obtained as t=−0.45 with P−value=0.661 for 12 degrees of freedom.

Answer 23

Section 2:

Expert Solution

Answer 24

Section 2:

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in peak heart rate of elite players and a university squad against the alternative that there is difference in peak heart rate of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where,

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(192.0−193.0)−06216+6213=−12.24=−0.45

The P-value for the provided one-sided test is P(T≥−0.45). For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t- statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=−0.45 lies beyond 0.25. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 1

Hence, the P-value is obtained as 0.661.

Answer 27

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Answer 28

To determine

To explain: The conclusion of the performed significance test.

Answer 29

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is no significant difference between the peak heart rate of the elite players of Canadian National team and a university squad.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The t-test statistic is obtained as -0.45 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.661 by using Excel in the previous part. This shows that P>0.05. So, the null hypothesis is not rejected. Hence, it can be concluded that the data strongly suggest that there is no significant difference between the peak heart rate of both types of players.

Answer 34

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

Answer 35

Section 3:

To determine

To test: Significance test to test whether there is difference in intermittent recovery test of Elite players from the Canadian National team and university squad.

Answer 36

Section 3:

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad. The t- test statistic is obtained as t=6.35 with P−value=0.000 for 12 degrees of freedom.

Answer 37

Section 3:

Expert Solution

Answer 38

Section 3:

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation: The hypothesis is considered as that there is no difference in intermittent recovery test of elite players and a university squad against the alternative that there is difference in intermittent recovery test of the elite players and university squad. Hence, the hypotheses are formulated as:

H0:μE=μUHa:μE≠μU

The two-sample t- test statistic is defined as:

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2

Where

x¯1=Mean of the treatment groupx¯2=Mean of the control groups1=Standard deviation of treatment groups2=Standard deviation of control group

n1=Size of treatment groupn2=Size of control groupμ1−μ2=Difference of means

The difference of means is considered as 0 points as the null hypothesis states that there is no difference between the two sets of players. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x1¯−x2¯)−(μ1−μ2)s12n1+s22n2=(1160−781)−0191216+129213=37959.67=6.35

For the second approximation, the degrees of freedom k is the smaller of n1−1 and n2−1.

n1−1 =16−1=15

n2−1 =13−1=12

So, the degree of freedom is 12. Compare the obtained value of t -statistic with the values in the table D for 12 degrees of freedom. The table D shows that the value of t=6.35 lies beyond the table. Use the formula =TDIST (t, d.f, tails) to determine the exact P- value in Excel software. The screenshot is provided below:

Introduction to the Practice of Statistics, Chapter 7.2, Problem 58E , additional homework tip 2

Therefore, the P-value is obtained as 0.000.

Answer 41

To determine

To explain: The conclusion of the performed significance test.

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.

Answer 42

To determine

To explain: The conclusion of the performed significance test.

Answer 43

Expert Solution

Answer to Problem 58E

Solution: The data strongly suggests that there is a significant difference between the intermittent recovery test of the elite players of Canadian National team and university squad.

Answer 44

Expert Solution

Answer 45

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The t-test statistic is obtained as 6.35 in the previous part. For 12 degrees of freedom, the P-value is obtained as 0.000 by using Excel in the previous part. This shows that P<0.05. So, the null hypothesis is rejected. Hence, it can be concluded that the data strongly suggests that there is a significant difference between the intermittent recovery tests of both the types of players.