Obtain the null and the alternative hypotheses.

Question

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Answer 1

Question

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

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

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iid 1. The CLT provides an approximate sampling distribution for the arithmetic average Ỹ of a random sample Y₁, . . ., Yn f(y). The parameters of the approximate sampling distribution depend on the mean and variance of the underlying random variables (i.e., the population mean and variance). The approximation can be written to emphasize this, using the expec- tation and variance of one of the random variables in the sample instead of the parameters μ, 02: YNEY, · (1 (EY,, varyi n For the following population distributions f, write the approximate distribution of the sample mean. (a) Exponential with rate ẞ: f(y) = ß exp{−ßy} 1 (b) Chi-square with degrees of freedom: f(y) = ( 4 ) 2 y = exp { — ½/ } г( (c) Poisson with rate λ: P(Y = y) = exp(-\} > y! y²

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Answer 2

Question

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

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

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

Answer 3

Question

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

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

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

Answer 4

Question

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

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

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 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

Answer 6

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

Answer 7

Chapter 12, Problem 11E

a.

To determine

Obtain the null and the alternative hypotheses.

Answer 8

a.

Expert Solution

Explanation of Solution

The null and alternative hypotheses are given below:

Null Hypothesis

H0:μ1=μ2=μ3

That is, the mean of all treatments are equal.

Alternative Hypothesis

H1: At least one treatment mean is different from others.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

Give the decision rule.

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

Answer 13

b.

To determine

Give the decision rule.

Answer 14

b.

Expert Solution

Explanation of Solution

The treatment and error degrees of freedom are given below:

Treatment degrees of freedom:

Degrees of freedom = k−1=3−1=2

Error degrees of freedom:

Degrees of freedom = n−k=12−3=9

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the critical F value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability.
Choose probability> F-distribution> calculate F given probability.
Enter P as 0.05.
Enter df1 as 2.
Enter df2 as 9.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 1

From the output, the critical F value is 4.26.

Decision rule:

If F>4.26, then reject the null hypothesis.

Therefore, the decision rule is to reject H0 if the computed value of F exceeds 4.26.

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

c.

To determine

Find the values of SST, SSE and SS total.

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

Answer 19

c.

To determine

Find the values of SST, SSE and SS total.

Answer 20

c.

Expert Solution

Answer to Problem 11E

The value of SST is 107.20.

The value of SSE is 9.47.

The value of SS total is 116.67.

Answer 21

c.

Expert Solution

Answer 22

c.

Expert Solution

Answer 23

Expert Solution

Answer 24

Explanation of Solution

Here, the level of significance (α) is 0.05.

Step-by-step procedure to obtain the sum of square total, sum of square treatment and sum of square error using Excel-MegaStat:

Choose MegStat > Analysis of Variance > One-Factor ANOVA.
Select the column of Treatment 1, Treatment 2 and Treatment 3 in Input range.
Click OK.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 2

From the output, the values of SST is 107.20, SSE is 9.47 and SS total is 116.67.

Answer 25

d.

To determine

Find an ANOVA table.

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

Answer 26

d.

To determine

Find an ANOVA table.

Answer 27

d.

Expert Solution

Explanation of Solution

From the output in Part (c), the ANOVA table is obtained.

The ANOVA table is given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	107.2	2	53.6	50.96
Error	9.47	9	1.05
Total	116.67	11

Answer 28

d.

Expert Solution

Answer 29

d.

Expert Solution

Answer 30

Expert Solution

Answer 31

e.

To determine

Find the decision regarding the null hypothesis.

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

Answer 32

e.

To determine

Find the decision regarding the null hypothesis.

Answer 33

e.

Expert Solution

Explanation of Solution

Conclusion:

The F value is 50.96 and the F critical value is 4.26.

Here, F value is greater than F critical value. That is, 50.96 > 4.26.

Using rejection rule, reject the null hypothesis.

Therefore, there is sufficient evidence that at least one mean of all treatment is differ from others.

Answer 34

e.

Expert Solution

Answer 35

e.

Expert Solution

Answer 36

Expert Solution

Answer 37

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.

Answer 38

f.

To determine

Check whether there is significant difference between treatment 1 and treatment 2, if null hypothesis is rejected by using the 95% level of confidence.

Answer 39

f.

Expert Solution

Explanation of Solution

A 95% confidence interval is as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)

Where,

x¯1 is the mean of treatment 1, x¯2 is the mean of treatment 2, n1 is the number of observations in the treatment 1, n2 is the number of observations in the treatment 2, t is the critical value with (n−k=9) degrees of freedom and MSE is mean square term obtained from the ANOVA table.

From the output in Part (c), the mean of treatment 1 is 9.7, mean of treatment 2 is 2.2, and MSE is 1.052.

Step-by-step procedure to obtain t-critical value using Excel-MegaStat:

In EXCEL, Select Add-Ins > MegaStat > Probability > t-Distribution.
Select calculate t given P.
Enter probability as 0.05.
Enter df as 9.
Under Shading, choose two-tail.
Click Ok.

Output using the Excel-MegaStat software is given below:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 12, Problem 11E , additional homework tip 3

From the output, the t is ±2.26. Now the 95% confidence interval is calculated as follows:

(Confidence interval for the difference in treatment means)=(x¯1−x¯2)±tMSE(1n1+1n2)=(9.7−2.2)±2.261.052(13+15)=7.5±(2.26×0.75)=7.5±1.70

=[7.5−1.70, 7.5+1.70]=[5.8, 9.2]

Therefore, a 95% confidence interval for that difference is 5.8 and 9.2. Here, 0 does not include in the confidence interval.

It means that there is a significant difference between the means of treatment 1 and treatment 2 because the endpoints have same sign or does not include zero.