Cola Weights Data Set 26 “Cola Weights and Volumes” in Appendix B lists the weights (lb) of the contents of cans of cola from four different samples: (1) regular Coke, (2) Diet Coke, (3) regular Pepsi, and (4) Diet Pepsi. The results from analysis of variance are shown on the top of the next page. What is the null hypothesis for this analysis of variance test? Based on the displayed results, what should you conclude about H 0 ? What do you conclude about equality of the mean weights from the four samples?

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Answer 1

Textbook Question

Chapter 12, Problem 1CQQ

Cola Weights Data Set 26 “Cola Weights and Volumes” in Appendix B lists the weights (lb) of the contents of cans of cola from four different samples: (1) regular Coke, (2) Diet Coke, (3) regular Pepsi, and (4) Diet Pepsi. The results from analysis of variance are shown on the top of the next page. What is the null hypothesis for this analysis of variance test? Based on the displayed results, what should you conclude about H₀? What do you conclude about equality of the mean weights from the four samples?

Chapter 12, Problem 1CQQ, Cola Weights Data Set 26 Cola Weights and Volumes in Appendix B lists the weights (lb) of the

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 + ...

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

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Answer 2

Textbook Question

Chapter 12, Problem 1CQQ

Cola Weights Data Set 26 “Cola Weights and Volumes” in Appendix B lists the weights (lb) of the contents of cans of cola from four different samples: (1) regular Coke, (2) Diet Coke, (3) regular Pepsi, and (4) Diet Pepsi. The results from analysis of variance are shown on the top of the next page. What is the null hypothesis for this analysis of variance test? Based on the displayed results, what should you conclude about H₀? What do you conclude about equality of the mean weights from the four samples?

Chapter 12, Problem 1CQQ, Cola Weights Data Set 26 Cola Weights and Volumes in Appendix B lists the weights (lb) of the

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 + ...

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

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Answer 3

Textbook Question

Chapter 12, Problem 1CQQ

Cola Weights Data Set 26 “Cola Weights and Volumes” in Appendix B lists the weights (lb) of the contents of cans of cola from four different samples: (1) regular Coke, (2) Diet Coke, (3) regular Pepsi, and (4) Diet Pepsi. The results from analysis of variance are shown on the top of the next page. What is the null hypothesis for this analysis of variance test? Based on the displayed results, what should you conclude about H₀? What do you conclude about equality of the mean weights from the four samples?

Chapter 12, Problem 1CQQ, Cola Weights Data Set 26 Cola Weights and Volumes in Appendix B lists the weights (lb) of the

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 + ...

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

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Answer 4

Textbook Question

Chapter 12, Problem 1CQQ

Cola Weights Data Set 26 “Cola Weights and Volumes” in Appendix B lists the weights (lb) of the contents of cans of cola from four different samples: (1) regular Coke, (2) Diet Coke, (3) regular Pepsi, and (4) Diet Pepsi. The results from analysis of variance are shown on the top of the next page. What is the null hypothesis for this analysis of variance test? Based on the displayed results, what should you conclude about H₀? What do you conclude about equality of the mean weights from the four samples?

Chapter 12, Problem 1CQQ, Cola Weights Data Set 26 Cola Weights and Volumes in Appendix B lists the weights (lb) of the

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 + ...

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

Answer 8

Expert Solution & Answer

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To obtain: The null hypothesis for the analysis of variance test.

To explain: The conclusion about H0.

To conclude: The equality of mean weights from the four samples.

Answer 12

Answer to Problem 1CQQ

The null hypothesis is H0:μ1=μ2=μ3=μ4.

The conclusion about H0 is that there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola from four different samples are the same.

No, the equality of mean weights from the four samples cannot be concluded.

Answer 13

Explanation of Solution

Given info:

The data shows that the weights of the contents of cans of cola from four different samples.

Calculation:

State the test hypotheses.

Let μ1 be the mean weight of regular Coke, μ2 be the mean weight of diet Coke and μ3 be the mean weights of regular Pepsi and μ4 be the mean weights of diet Pepsi.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

H1: At least one of the means is different from the other means

From the Minitab output,

The test statistic F is 503.06 and the P-value is 0.000.

Decision criteria for the P-value method:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is lesser than the significance level.

That is, 0.000(=P-value)<0.05(=α).

Here, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that the mean weights of the contents of cans of cola for four different samples are same.

Justification:

No, the equality of mean weights from the four samples cannot be concluded because the mean weights of the contents of cans of cola from four different samples are different

Answer 14

Solution Summary: The author explains the null hypothesis for the analysis of variance test.

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Answer to Problem 1CQQ

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Chapter 12 Solutions