Question

A medical researcher believes that a drug changes the body's temperature. Seven test subjects are randomly selected and the body temperature of each is measured. The subjects are then given the drug, and after 30 minutes, the body temperature of each is measured again. The results are listed in the table below. Is there enough evidence to conclude that the drug changes the body's temperature?

Let d=(body temperature after taking drug)−(body temperature before taking drug)d=(body temperature after taking drug)−(body temperature before taking drug). Use a significance level of α=0.02 for the test. Assume that the body temperatures are normally distributed for the population of people both before and after taking the drug.

Subject	1	2	3	4	5	6	7
Temperature (before)	99.6	98.6	99.7	99.6	99.9	100.3	100.6
Temperature (after)	98.9	97.8	100.2	99.4	99.1	100.1	100.3

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5:Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places

Step 4 of 5:Determine the decision rule for rejecting the null hypothesis H 0   . Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test. (Fail to reject or reject null)

Answer 1

Step 1 of 5:
H0: Null Hypothesis: =0 ( drug does not change the body's temperature )

HA: Alternative Hypothesis:0 ( drug changes the body's temperature ) (Claim)

Step 2 of 5:

From the given data, values of d = Temperature after - Temperature before are calculated as follows:

d = Temperature after - Temperature before = - 0.7, - 0.8, 0.5, - 0.2, -0.8, - 0.2, - 0.3

From d values,the following statistics are calculated:

n= 7

= - 0.357

s_d = 0.465 ( Round your answer to one decimal place. 0.5)

the value of the standard deviation of the paired differences = 0.5

Step 3 of 5:

Test Statistic is given by:

the value of the test statistic = - 2.032

Step 4 of 5:

df = 7 - 1 = 6

= 0.02

From Table, critical values of t = 3.143

Decision Rule:

Reject H0 if t < - 3.143 OR t > 3.143

Step 5 of5:

Since calculated value of t = - 2.032 is greater than critical value of t = - 3.143:

the difference is not significant. Fail to reject null hypothesis.