Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
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Chapter 9, Problem 1CS
To determine

State the P–value for each test by performing a test of the null hypothesis that the group means or group proportions are equal versus the alternative that they are not equal.

Expert Solution & Answer
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Answer to Problem 1CS

The P–value for the each test is as follows:

CharacteristicP–value
Age0.057
Systolic blood pressure0.000
Diastolic blood pressure0.037
Treatment for hypertension0.819
Atrial fibrillation0.337
Diabetes0.875
Cigarette smoking0.343
Coronary bypass surgery0.762

Explanation of Solution

Calculation:

The data represents the means and standard deviations corresponding to three characteristics of standard treatment and new treatment.

Furthermore, the data corresponding to the percentage with the characteristics for the 5 characteristics for standard treatment and new treatment is given.

Here, it is given that the sample size of standard treatment is n1=731 and the sample size of new treatment is n2=1,089.

Let x¯i's be the sample means of each characteristics, si's are the sample standard deviations of each characteristics and μi's be the population means of each characteristics.

Hypothesis test for the variable “Age”:

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

That is, there is no significant difference between the mean age of new treatment and standard treatment.

Alternate hypothesis:

H1:μ1μ2

That is, there is a significant difference between the mean age of new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populations means using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first sample, enter the sample size as 731, sample mean as 64 and sample standard deviation as 11.
  • In second sample, enter the sample size as 1,089, sample mean as 65 and sample standard deviation as 11.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  1

From, the MINITAB output the P-value corresponding to the variable “Age” is 0.057.

Hypothesis test for the variable “Systolic blood pressure”:

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

That is, there is no significant difference between the mean systolic blood pressure of new treatment and standard treatment.

Alternate hypothesis:

H1:μ1μ2

That is, there is a significant difference between the mean systolic blood pressure of new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two population means using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first sample, enter the sample size as 731, sample mean as 121 and sample standard deviation as 18.
  • In second sample, enter the sample size as 1,089, sample mean as 124 and sample standard deviation as 17.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  2

From, the MINITAB output the P-value corresponding to the variable “Systolic blood pressure” is 0.000.

Hypothesis test for the variable “Diastolic blood pressure”:

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2

That is, there is no significant difference between the mean diastolic blood pressure of new treatment and standard treatment.

Alternate hypothesis:

H1:μ1μ2

That is, there is a significant difference between the mean diastolic blood pressure of new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populations means using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first sample, enter the sample size as 731, sample mean as 71 and sample standard deviation as 10.
  • In second sample, enter the sample size as 1,089, sample mean as 72 and sample standard deviation as 10.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  3

From, the MINITAB output the P-value corresponding to the variable “Diastolic blood pressure” is 0.037.

Hypothesis test for the variable “Treatment for hypertension”:

The sample size of standard treatment is n1=731. The percentage with the characteristics “Treatment for hypertension under standard treatment” is p1=0.632.

Hence, the specified characteristics of treatment for hypertension under new treatment is x1=n1×p1=461.992.

The sample size of new treatment is n2=1,089. The percentage with the characteristics “Treatment for hypertension under new treatment” is p2=0.637.

Hence, the specified characteristics of treatment for hypertension under new treatment is x2=n2×p2=693.693.

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.

Alternate hypothesis:

H1:p1p2

That is, there is a significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two population proportions using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Proportions.
  • Choose Summarized data.
  • In first sample, enter the number of trials as 731 and number of events as 462.
  • In second sample, enter the number of trials as 1,089 and number of events as 694.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  4

From, the MINITAB output the P-value corresponding to the variable “Treatment for hyper tension” is 0.819.

Hypothesis test for the variable “Atrial fibrillation”:

The sample size of standard treatment is n1=731. The percentage with the characteristics “Atrial fibrillation under standard treatment” is p1=0.126.

Hence, the specified characteristics of treatment for Atrial fibrillation under standard treatment is x1=n1×p1=92.106.

The sample size of new treatment is n2=1,089. The percentage with the characteristics “Atrial fibrillation under new treatment” is p2=0.111.

Hence, the specified characteristics of Atrial fibrillation under new treatment is x2=n2×p2=120.879.

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.

Alternate hypothesis:

H1:p1p2

That is, there is a significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populationproportions using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Proportions.
  • Choose Summarized data.
  • In first sample, enter the number of trials as 731 and number of events as 92.
  • In second sample, enter the number of trials as 1,089 and number of events as 121.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  5

From, the MINITAB output the P-value corresponding to the variable “Atrial fibrillation” is 0.337.

Hypothesis test for the variable “Diabetes”:

The sample size of standard treatment is n1=731. The percentage with the characteristics “Diabetes under standard treatment” is p1=0.302.

Hence, the specified characteristics of diabetes under standard treatment is x1=n1×p1=220.762.

The sample size of new treatment is n2=1,089. The percentage with the characteristics “Diabetes under new treatment” is p2=0.306.

Hence, the specified characteristics of diabetes under new treatment is x2=n2×p2=333.234.

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of Diabetes under new treatment and standard treatment.

Alternate hypothesis:

H1:p1p2

That is, there is a significant difference between the proportion of Diabetes under new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populationproportions using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Proportions.
  • Choose Summarized data.
  • In first sample, enter the number of trials as 731 and number of events as 221.
  • In second sample, enter the number of trials as 1,089 and number of events as 333.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  6

From, the MINITAB output the P-value corresponding to the variable “Diabetes” is 0.875.

Hypothesis test for the variable “Cigarette smoking”:

The sample size of standard treatment is n1=731. The percentage with the characteristics “Cigarette smoking under standard treatment” is p1=0.128.

Hence, the specified characteristics of cigarette smoking under standard treatment is x1=n1×p1=93.568.

The sample size of new treatment is n2=1,089. The percentage with the characteristics “Cigarette smoking under new treatment” is p2=0.114.

Hence, the specified characteristics of cigarette smoking under new treatment is x2=n2×p2=124.146.

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of Cigarette smoking under new treatment and standard treatment.

Alternate hypothesis:

H1:p1p2

That is, there is a significant difference between the proportion of Cigarette smoking under new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populationsproportions using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Proportions.
  • Choose Summarized data.
  • In first sample, enter the number of trials as 731 and number of events as 94.
  • In second sample, enter the number of trials as 1,089 and number of events as 124.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  7

From, the MINITAB output the P-value corresponding to the variable “Cigarette smoking” is 0.343.

Hypothesis test for the variable “Coronary bypass surgery”:

The sample size of standard treatment is n1=731. The percentage with the characteristics “Coronary bypass surgery under standard treatment” is p1=0.285.

Hence, the specified characteristics of Coronary bypass surgery under standard treatment is x1=n1×p1=208.335.

The sample size of new treatment is n2=1,089. The percentage with the characteristics “Coronary bypass surgery under new treatment” is p2=0.291.

Hence, the specified characteristics of coronary bypass surgery under new treatment is x2=n2×p2=316.899.

The hypotheses are given below:

Null hypothesis:

H0:p1=p2

That is, there is no significant difference between the proportion of Coronary bypass surgery under new treatment and standard treatment.

Alternate hypothesis:

H1:p1p2

That is, there is a significant difference between the proportion of Coronary bypass surgery under new treatment and standard treatment.

P-value:

Software Procedure:

Step by step procedure to perform hypothesis test for the difference between the two populationproportions using the MINITAB software:

  • Choose Stat > Basic Statistics > 2-Proportions.
  • Choose Summarized data.
  • In first sample, enter the number of trials as 1,089 and number of events as 317.
  • In second sample, enter the number of trials as 731 and number of events as 208.
  • Choose Options.
  • In Confidence level, enter 95.
  • Enter 0 in Hypothesized value.
  • In Alternative, select not equal to
  • Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9, Problem 1CS , additional homework tip  8

From, the MINITAB output the P-value corresponding to the variable “Coronary bypass surgery” is 0.762.

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Chapter 9 Solutions

Essential Statistics

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