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). Use a significance level of a = 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 Copy Data

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: Find the p-value for the hypothesis test. Round your answer to four decimal places.

Step 5 of 5: Draw a conclusion for the hypothesis test.

Answer 1

The table given below ,

Subject	Before(X)	After(Y)	di=Y-X	di^2
1	99.6	98.9	-0.7	0.49
2	98.6	97.8	-0.8	0.64
3	99.7	100.2	0.5	0.25
4	99.6	99.4	-0.2	0.04
5	99.9	99.1	-0.8	0.64
6	100.3	100.1	-0.2	0.04
7	100.6	100.3	-0.3	0.09
		Sum	-2.5	2.19

From table ,

$\bar{d}=\frac{\sum d_i}{n}=\frac{-2.5}{7}=-0.3571$

Part 1 of 5 : The null and alternative hypothesis is ,

$H_0:\mu_d=0$

$H_a:\mu_d\neq 0$

The test is two-tailed test.

Part 2 of 5 : The value of the standard deviation of the paired differences is ,

$s_d=\sqrt{\frac{\sum d_i^2-n*\bar{d}^2}{n-1}}=\sqrt{\frac{2.19-7*(-0.3571)^2}{7-1}}=0.46$

Part 3 of 5 : The value of the test statistic is ,

$t_{stat}=\frac{\bar{d}}{s_d/\sqrt{n}}=\frac{-0.3571}{0.46/\sqrt{7}}=-2.054$

Part 4 of 5 : Now , df=degrees of freedom=n-1=7-1=6

The p-value is ,

p-value=$P(t_{df}>|t_{stat}|)=P(t_{6}>2.054)=0.0429$ ; The Excel function is , =TDIST(2.054,6,1)

Part 5 of 5 : Decision : Here , p-value >0.02

Therefore , fail to reject Ho.

Conclusion : There is not enough evidence to conclude that the drug changes the body temperature.