Question

Suppose that we have developed a new drug for critically ill patients. In the control group of 36 patients, given a placebo, the average time to release from the ICU was 12.8 days. In the group of 50 patients that we gave our experimental drug, the average time to release from the ICU was 12.1 days. Assuming both populations had the same standard deviation of 3.6 days and the same proportion of recoveries, would you say there is evidence of the drug's effectiveness at a 5% significance level? SHOW WORK!!

The alternative hypothesis for this test is :  

The distribution (choose from Z, T, X^2, or F) for this test is  

The critical value for this test is

The test value for this test is

Based on this information, we (choose reject or fail to reject)   the null hypothesis

Based on this we (choose from do or do not)   have evidence that this drug is effective.

Answer 1

using excel>addin>phstat>two sample test

we have

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.05
Population 1 Sample
Sample Size	36
Sample Mean	12.8
Sample Standard Deviation	3.6
Population 2 Sample
Sample Size	50
Sample Mean	12.1
Sample Standard Deviation	3.6
Intermediate Calculations
Population 1 Sample Degrees of Freedom	35
Population 2 Sample Degrees of Freedom	49
Total Degrees of Freedom	84
Pooled Variance	12.9600
Standard Error	0.7869
Difference in Sample Means	0.7000
t Test Statistic	0.8896
Upper-Tail Test
Upper Critical Value	1.6632
p-Value	0.1881
Do not reject the null hypothesis

The distribution for this test is t distributionThe alternative hypothesis for this test is : Ha: u₁ > u₂  

The critical value for this test is 1.6632

The test value for this test is 0.8896

Based on this information, we fail to reject the null hypothesis

Based on this we do not have evidence that this drug is effective.