For the preceding problem you should find that there are significant differences among the three treatments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 19 but doubled the SS values within each sample. a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 19. b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained m problem 19? c. Use an A VOVA with α – .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in purl b?)

Question

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Answer 1

Textbook Question

Chapter 12, Problem 20P

For the preceding problem you should find that there are significant differences among the three treatments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 19 but doubled the SS values within each sample.

Chapter 12, Problem 20P, For the preceding problem you should find that there are significant differences among the three

a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 19.
b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained m problem 19?
c. Use an A VOVA with α – .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in purl b?)

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 + ...

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

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Answer 2

Textbook Question

Chapter 12, Problem 20P

For the preceding problem you should find that there are significant differences among the three treatments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 19 but doubled the SS values within each sample.

Chapter 12, Problem 20P, For the preceding problem you should find that there are significant differences among the three

a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 19.
b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained m problem 19?
c. Use an A VOVA with α – .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in purl b?)

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 + ...

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

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Answer 3

Textbook Question

Chapter 12, Problem 20P

For the preceding problem you should find that there are significant differences among the three treatments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 19 but doubled the SS values within each sample.

Chapter 12, Problem 20P, For the preceding problem you should find that there are significant differences among the three

a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 19.
b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained m problem 19?
c. Use an A VOVA with α – .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in purl b?)

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 + ...

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

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Answer 4

Textbook Question

Chapter 12, Problem 20P

For the preceding problem you should find that there are significant differences among the three treatments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 19 but doubled the SS values within each sample.

Chapter 12, Problem 20P, For the preceding problem you should find that there are significant differences among the three

a. Calculate the sample variance for each of the three samples. Describe how these sample variances compare with those from problem 19.
b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained m problem 19?
c. Use an A VOVA with α – .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in purl b?)

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 + ...

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

Answer 8

a.

Expert Solution

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Variances for each sample.

To compare: Obtained sample variances with sample variances from problem 19.

Answer 13

Answer to Problem 20P

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Answer 14

Explanation of Solution

Given info:

The following values are from an independent-measures study comparing three treatment conditions.

Treatment
I	II	III
n=6	n=6	n=6	N=18
M=1	M=2	M=6	G=54
SS=120	SS=130	SS=80	∑X2=576

Calculation:

Let k represents the numbers of treatments. Then,

k=3

For treatment 1

n=6SS=120

Let S12 is the sample variance for treatment 1. Then,

S12=SSn−1=1206−1=1205=24

For treatment 2

n=6SS=130

Let S22 is the sample variance for treatment 2. Then,

S22=SSn−1=1306−1=1305=26

For treatment 3

n=6SS=80

Let S32 is the sample variance for treatment 3. Then,

S32=SSn−1=806−1=805=16

The sample variances for treatment I, treatment II and treatment III in problem 19 are 12, 13 and 8 respectively. But sample variances for all treatments in this problem are twice of sample variances in problem 19.

Conclusion:

Sample variances for treatment 1, treatment 2, and treatment 3 are 24, 26 and 16 respectively. Obtained sample variance for each treatment are twice of the sample variance in problem 19.

Answer 15

b.

Expert Solution

To determine

To predict: The outcome of the analysis.

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To predict: The outcome of the analysis.

Answer 20

Answer to Problem 20P

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Answer 21

Explanation of Solution

From part a. sample variance for each treatment are twice of the sample variances in problem 19.

The formula for F ratio is:

F=MSbetweenMSwithin

F-ratio is inversely proportional to the mean sum of squares within treatments. So, large sample variances give large mean sum of squares within treatments and hence small value of F-ratio. Therefore, on increasing the size of sample variances, the size of F-ratio would decreases. So, when sample variances increases, chances to reject the null hypothesis decreases.

Conclusion:

The size of the F-ratio decreases and there are fewer chances to reject the null hypothesis.

Answer 22

c.

Expert Solution

To determine

There are any significant differences among the three treatment means.

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

There are any significant differences among the three treatment means.

Answer 27

Answer to Problem 20P

There are no significant mean differences between the three treatments.

Answer 28

Explanation of Solution

Calculation:

Let k represents the numbers of treatments. Then,

k=3

Then, degrees of freedom corresponding to SSbetween are:

dfbetween=k−1=3−1=2

Then, degrees of freedom corresponding to SSwithin are:

dfwithin=N−k=18−3=15

SSwithin is calculated as:

SSwithin=∑SS=120+130+80=330

Total sum of squares is given as:

SStotal=∑X2−G2N=576−(54)218=576−162=414

Sum of squares between treatments is given as:

SSbetween=SStotal−SSwithin=414−330=84

F-ratio is calculated as:

F=SSbetweendfbetweenSSwithindfwithin=84233015=4222=1.91

From the table B.4 of appendix B in the textbook, critical value corresponding to df=2,15 is 3.68.

Since, F-ratio is less than the critical value; therefore, we fail to reject the null hypothesis and conclude that there are no significant mean differences between the treatments.

Conclusion:

There are no significant mean differences between the treatments. Also, the answer agree with prediction in part b.