Question

Please state the step number next to the answer given, thank you!

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 before taking drug)−(body temperature after taking drug)d=(body temperature before taking drug)−(body temperature after taking drug). Use a significance level of α=0.05 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)	100.6	100	97.8	100.6	100.4	98.9	99.2
Temperature (after)	100.3	99.3	98.4	100.4	100.1	98.3	98.4

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 two decimal places.
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 H0. Round the numerical portion of your answer to three decimal places.
Step 5 of 5: Make the decision for the hypothesis test. (Reject or Fail to Reject)

Answer 1

Step 1

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

Step 2

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	100.6	100.3	0.3
	100	99.3	0.7
	97.8	98.4	-0.6
	100.6	100.4	0.2
	100.4	100.1	0.3
	98.9	98.3	0.6
	99.2	98.4	0.8
Average	99.643	99.314	0.329
St. Dev.	1.055	0.955	0.468
n	7	7	7

For the score differences, we have

calculation:

Sample variance is

Therefore, the sample standard deviation S_D is

Step 3

Test Statistics

The t-statistic is computed as shown in the following formula:

Step 4

Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=6.

Hence, it is found that the critical value for this two-tailed test is tc​=2.447, for α=0.05 and df=6.

The rejection region for this two-tailed test is

Step 5

The decision about the null hypothesis

Since it is observed that ∣t∣=1.857≤tc​=2.447, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1126, and since p=0.1126≥0.05, it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

Graphically

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