
Check whether the results obtained in Exercise-2 are same with the results obtained in Exercise-4.

Answer to Problem 5CS
Yes, the results obtained in Exercise-2 are same with the results obtained in Exercise-4.
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
Hypothesis test for the variable “Age”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the mean age of new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the mean age of new treatment and standard treatment.
Hypothesis test for the variable “Systolic blood pressure”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the mean systolic blood pressure of new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the mean systolic blood pressure of new treatment and standard treatment.
Hypothesis test for the variable “Diastolic blood pressure”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the meandiastolic blood pressure of new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the meandiastolic blood pressure of new treatment and standard treatment.
Hypothesis test for the variable “Treatment for hypertension”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.
Hypothesis test for the variable “Atrial fibrillation”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.
Hypothesis test for the variable “Diabetes”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the proportion of Diabetes under new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the proportion of Diabetes under new treatment and standard treatment.
Hypothesis test for the variable “Cigarette smoking”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the proportion of Cigarette smoking under new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the proportion of Cigarette smoking under new treatment and standard treatment.
Hypothesis test for the variable “Coronary bypass surgery”:
The hypotheses are given below:
Null hypothesis:
That is, there is no significant difference between the proportion of Coronary bypass surgery under new treatment and standard treatment.
Alternate hypothesis:
That is, there is a significant difference between the proportion of Coronary bypass surgery under new treatment and standard treatment.
The results obtained in Exercise-2 are as follows:
Decision rule based on P-value:
If
If
Characteristic | P–value |
Relation with P–value and |
Acceptance of rejection of null hypothesis |
Age | 0.057 | Fail to reject the null hypothesis | |
Systolic blood pressure | 0.000 |
Reject the null hypothesis | |
Diastolic blood pressure | 0.037 |
Reject the null hypothesis | |
Treatment for hypertension | 0.819 | Fail to reject the null hypothesis | |
Atrial fibrillation | 0.337 | Fail to reject the null hypothesis | |
Diabetes | 0.875 | Fail to reject the null hypothesis | |
Cigarette smoking | 0.343 | Fail to reject the null hypothesis | |
Coronary bypass surgery | 0.762 | Fail to reject the null hypothesis |
Conclusions:
There is no significant difference between the mean age of new treatment and standard treatment.
There is a significant difference between the mean systolic blood pressure of new treatment and standard treatment.
There is a significant difference between the mean diastolic blood pressure of new treatment and standard treatment.
There is no significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.
There is no significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.
There is no significant difference between the proportion of cigarette smoking under new treatment and standard treatment.
There is no significant difference between the proportion of coronary bypass surgery under new treatment and standard treatment.
From this it is known that, the assignments to the groups are not balanced for the health characteristics “Systolic blood pressure” and “Diastolic blood pressure”.
Thus, the assignments to the groups are not balanced for the health characteristics “Systolic blood pressure” and “Diastolic blood pressure”.
The results obtained in Exercise-4 are as follows:
Decision in the given scenario:
If “0” lies between the confidence interval, then fail to reject the null hypothesis
If “0” does not lie between the confidence interval, then reject the null hypothesis
Conclusion based on confidence intervals:
Characteristic | P–value | Whether or not “0” lies in the confidence interval |
Acceptance of rejection of null hypothesis |
Age |
“0” lies in the confidence interval | Fail to reject the null hypothesis | |
Systolic blood pressure |
“0” does not lie in the confidence interval |
Reject the null hypothesis | |
Diastolic blood pressure |
“0” does not lie in the confidence interval |
Reject the null hypothesis | |
Treatment for hypertension |
“0” lies in the confidence interval | Fail to reject the null hypothesis | |
Atrial fibrillation |
“0” lies in the confidence interval | Fail to reject the null hypothesis | |
Diabetes |
“0” lies in the confidence interval | Fail to reject the null hypothesis | |
Cigarette smoking |
“0” lies in the confidence interval | Fail to reject the null hypothesis | |
Coronary bypass surgery |
“0” lies in the confidence interval | Fail to reject the null hypothesis |
Conclusions:
There is no significant difference between the mean age of new treatment and standard treatment.
There is a significant difference between the mean systolic blood pressure of new treatment and standard treatment.
There is a significant difference between the mean diastolic blood pressure of new treatment and standard treatment.
There is no significant difference between the proportion of treatment for hypertension under new treatment and standard treatment.
There is no significant difference between the proportion of Atrial fibrillation under new treatment and standard treatment.
There is no significant difference between the proportion of cigarette smoking under new treatment and standard treatment.
There is no significant difference between the proportion of coronary bypass surgery under new treatment and standard treatment.
From this it is known that, the assignments to the groups are not balanced for the health characteristics “Systolic blood pressure” and “Diastolic blood pressure”.
Thus, the assignments to the groups are not balanced for the health characteristics “Systolic blood pressure” and “Diastolic blood pressure”.
From the above conclusions, it is clear that that the results obtained in Exercise-2 are same with the results obtained in Exercise-4.
The close relationship between confidence interval and testing of hypothesis sates that, if the value of the population parameter mentioned in the null hypothesis is contained in the 95% interval then the null hypothesis cannot be rejected at 5% level. If the value specified by the null hypothesis is not in the interval then the null hypothesis can be rejected at 5% level.
Thus, the results obtained in Exercise-2 are same with the results obtained in Exercise-4.
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Chapter 9 Solutions
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