Check whether there is any difference in the mean ozone level among the four cities at the α = 0.05 level of significance.

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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 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

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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 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

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Answer 3

Question

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

Chapter 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

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Answer 4

Question

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

Chapter 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

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

Answer 5

Chapter 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer 6

Chapter 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer 7

Chapter 13, Problem 36SE

To determine

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer 8

Expert Solution & Answer

Answer to Problem 36SE

There is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

The data represent the levels of ozone pollution in four cities for 10 dates from last year.

Denote treatment means as city and block means as dates.

State the hypotheses:

For treatments

Null hypothesis:

H0T: There is no significant difference among the treatment means.

Alternative hypothesis:

HaT: There is a significant difference among the treatment means.

For Blocks

Null hypothesis:

H0B: There is no significant difference among the block means.

Alternative hypothesis:

HaB: There is a significant difference among the block means.

There are 4 cities (Treatments) and 10 dates from which the data was collected (Blocks). The level of significance is 0.05.

Randomized block design:

Software procedure:

Step-by-step procedure to obtain the randomized block design using the EXCEL:

Open new EXCEL worksheet.
Enter the data values in the EXCEL sheet.
On Data tab in Analysis group, click Data Analysis.
Select Anova: Two-Factor Without Replication.
Click Ok.
Click in the Input Range box, select the range $A$1: $E$11.
Select Labels and Enter Alpha as 0.05.
Click in the Output range, select $G$1.
Click OK.

Output obtained using the EXCEL software is given below:

Modern Business Statistics with Microsoft Excel (MindTap Course List), Chapter 13, Problem 36SE

From the EXCEL output, the F-ratio for treatments (city) is 2.42 and the p-value is 0.088. The F-ratio for blocks (Date) is 4.55 and the p-value is 0.001.

Decision:

If p-value≤α,, reject the null hypothesis H0

If p-value>α, fail to reject the null hypothesis H0

Conclusion:

Here, the p-value of treatments is greater than the level of significance.

That is, p-value(=0.088)>α(=0.05).

Therefore, the null hypothesis of treatments is not rejected.

Thus, there is no sufficient evidence to conclude that there is a significant difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer 12

Ch. 13.2 - 1. The following data are from a completely...Ch. 13.2 - 2. In a completely randomized design, seven...Ch. 13.2 - 3. Refer to exercise 2. what hypotheses are...Ch. 13.2 - Prob. 4E Ch. 13.2 - 5. In a completely randomized design, 12...Ch. 13.2 - Prob. 6E Ch. 13.2 - 7. Three different methods for assembling a...Ch. 13.2 - Prob. 8E Ch. 13.2 - 9. To study the effect of temperature on yield in...Ch. 13.2 - Prob. 10E

Check whether there is any difference in the mean ozone level among the four cities at the α = 0.05 level of significance.

Check whether there is any difference in the mean ozone level among the four cities at the α=0.05 level of significance.

Answer to Problem 36SE

Explanation of Solution

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Chapter 13 Solutions