The article “Gypsum Effect on the Aggregate Size and Geometry of Three Sodic Soils Under Reclamation” (I. Lebron, D. Suarez, and T. Yoshida, Journal of the Soil Science Society of America , 2002:92–98) reports on an experiment in which gypsum was added in various amounts to soil samples before leaching. One of the outcomes of interest was the pH of the soil. Gypsum was added in four different amounts. Three soil samples received each amount added. The pH measurements of the samples are presented in the following table. Gypsum (g/kg) pH 0.00 7.88 7.72 7.68 0.11 7.81 7.64 7.85 0.19 7.84 7.63 7.87 0.38 7.80 7.73 8.00 Can you conclude that the pH differs with the amount of gypsum added? Provide the value of the test statistic and the P -value.

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Answer 1

Textbook Question

Chapter 9, Problem 1SE

The article “Gypsum Effect on the Aggregate Size and Geometry of Three Sodic Soils Under Reclamation” (I. Lebron, D. Suarez, and T. Yoshida, Journal of the Soil Science Society of America, 2002:92–98) reports on an experiment in which gypsum was added in various amounts to soil samples before leaching. One of the outcomes of interest was the pH of the soil. Gypsum was added in four different amounts. Three soil samples received each amount added. The pH measurements of the samples are presented in the following table.

Gypsum (g/kg)	pH
0.00	7.88	7.72	7.68
0.11	7.81	7.64	7.85
0.19	7.84	7.63	7.87
0.38	7.80	7.73	8.00

Can you conclude that the pH differs with the amount of gypsum added? Provide the value of the test statistic and the P-value.

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

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Answer 2

Textbook Question

Chapter 9, Problem 1SE

The article “Gypsum Effect on the Aggregate Size and Geometry of Three Sodic Soils Under Reclamation” (I. Lebron, D. Suarez, and T. Yoshida, Journal of the Soil Science Society of America, 2002:92–98) reports on an experiment in which gypsum was added in various amounts to soil samples before leaching. One of the outcomes of interest was the pH of the soil. Gypsum was added in four different amounts. Three soil samples received each amount added. The pH measurements of the samples are presented in the following table.

Gypsum (g/kg)	pH
0.00	7.88	7.72	7.68
0.11	7.81	7.64	7.85
0.19	7.84	7.63	7.87
0.38	7.80	7.73	8.00

Can you conclude that the pH differs with the amount of gypsum added? Provide the value of the test statistic and the P-value.

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

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Answer 3

Textbook Question

Chapter 9, Problem 1SE

The article “Gypsum Effect on the Aggregate Size and Geometry of Three Sodic Soils Under Reclamation” (I. Lebron, D. Suarez, and T. Yoshida, Journal of the Soil Science Society of America, 2002:92–98) reports on an experiment in which gypsum was added in various amounts to soil samples before leaching. One of the outcomes of interest was the pH of the soil. Gypsum was added in four different amounts. Three soil samples received each amount added. The pH measurements of the samples are presented in the following table.

Gypsum (g/kg)	pH
0.00	7.88	7.72	7.68
0.11	7.81	7.64	7.85
0.19	7.84	7.63	7.87
0.38	7.80	7.73	8.00

Can you conclude that the pH differs with the amount of gypsum added? Provide the value of the test statistic and the P-value.

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

Want to see more full solutions like this?

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

Answer 4

Textbook Question

Chapter 9, Problem 1SE

The article “Gypsum Effect on the Aggregate Size and Geometry of Three Sodic Soils Under Reclamation” (I. Lebron, D. Suarez, and T. Yoshida, Journal of the Soil Science Society of America, 2002:92–98) reports on an experiment in which gypsum was added in various amounts to soil samples before leaching. One of the outcomes of interest was the pH of the soil. Gypsum was added in four different amounts. Three soil samples received each amount added. The pH measurements of the samples are presented in the following table.

Gypsum (g/kg)	pH
0.00	7.88	7.72	7.68
0.11	7.81	7.64	7.85
0.19	7.84	7.63	7.87
0.38	7.80	7.73	8.00

Can you conclude that the pH differs with the amount of gypsum added? Provide the value of the test statistic and the P-value.

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

Answer 8

Expert Solution & Answer

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Check whether the mean pH level differs for the different amounts of gypsum added.

Find the value of test statistic and P-value.

Answer 12

Answer to Problem 1SE

The test statistic is 0.29 and the P-value is 0.832.

There is no sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added.

Answer 13

Explanation of Solution

Given info:

The design variable is the Gypsum and the response is the pH measurement. The table provides the pH measurement corresponding to Gypsum.

Calculation:

State the hypotheses:

Null hypothesis:

H0:μ1=μ2=μ3=μ4

Alternative hypothesis:

Ha: At least two of the μi′s differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the column of pH.
In Factor, enter the column of Gypsum.
In Confidence level, enter 0.95.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 9, Problem 1SE

From the ANOVA table, the P-value is 0.832 and the F-value is 0.29.

Decision:

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

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

Since, the level of significance is not specified; the prior level of significance α=0.05 can be used.

Conclusion:

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

That is, P-value(=0.832)>α(=0.05).

By rejection rule, fails to reject the null hypothesis.

There is sufficient evidence to conclude that the mean pH level differs for the different amounts of gypsum added at α=0.05 level of significance.

Answer 14

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