Question

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 18 patients were recorded before administering the drug and after. The average difference in cholesterol levels (after - before) was 4.19 mg/dL with a standard deviation of 8.055 mg/dL. Using this information, the calculated 90% confidence paired-t interval is (0.887, 7.493). Which of the following is the best interpretation of this interval?

Question 8 options:

1)	The proportion of all patients that had a difference in cholesterol levels between those on the drug and those who are not is 90%.

2)	We are 90% confident that the average difference in the cholesterol levels of the patients sampled is between 0.887 and 7.493.

3)	We are certain the average difference in cholesterol levels between those who would take the drug and those who would not is between 0.887 and 7.493.

4)	We are 90% confident that the difference between the average cholesterol level for those on the drug and the average cholesterol level for those not on the drug is between 0.887 and 7.493.

5)	We are 90% confident that the average difference in cholesterol levels between those who would take the drug and those who would not is between 0.887 and 7.493.

A medical researcher wants to determine if the average hospital stay after a certain procedure is less than 14.97 days. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 14.97, Alternative Hypothesis: μ < 14.97. If the researcher randomly samples 26 patients that underwent the procedure and determines their average hospital stay was 15.68 days with a standard deviation of 7.183 days, what is the test statistic and p-value of this test?

Question 9 options:

1)	Test Statistic: -0.504, P-Value: 0.6907

2)	Test Statistic: 0.504, P-Value: 0.6907

3)	Test Statistic: 0.504, P-Value: 0.3093

4)	Test Statistic: -0.504, P-Value: 0.3093

5)	Test Statistic: 0.504, P-Value: 1.3814

Answer 1

Question 1

We are 90% confident that the average difference in cholesterol levels between those who would take the drug and those who would not is between 0.887 and 7.493.

Question 2

Test Statistic :-


t = 0.504

Test Criteria :-
Reject null hypothesis if

= 0.504 > - 2.485
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 0.504 ) = 0.6907
Reject null hypothesis if P value < level of significance
P - value = 0.6907 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

Test Statistic: 0.504, P-Value: 0.6907