Question

A pharmaceutical company claims that its new drug reduces systolic blood pressure. The systolic blood pressure (in millimeters of mercury) for nine patients before taking the new drug and 2 hours after taking the drug are shown in the table below. Is there enough evidence to support the company's claim?

Let d=(blood pressure before taking new drug)−(blood pressure after taking new drug). Use a significance level of α=0.01 for the test. Assume that the systolic blood pressure levels are normally distributed for the population of patients both before and after taking the new drug.

Patient	1	2	3	4	5	6	7	8	9
Blood pressure (before)	169	163	182	194	172	193	147	154	175
Blood pressure (after)	162	144	162	178	146	176	140	140	150

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

Ho: μd (=,≠,<,>,≤,≥) 0

Ha: μd (=,≠,<,>,≤,≥) 0

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.

Reject Ho if (t, I t I) (<,>) _____

Step 5 of 5: Make the decision for the hypothesis test.

Reject Null Hypothesis

Fail to Reject Null Hypothesis

Answer 1

d=(blood pressure before taking new drug)−(blood pressure after taking new drug)

company claim:  its new drug reduces systolic blood pressure :

Step 1 of 5

Null hypothesis H_o :

Alternate hypothesis : H_a

Right tailed test:

Step 2 of 5

Sample mean of  paired differences :

n: sample size = 9

value of the standard deviation of the paired differences : s_d

Patient	Blood pressure (before)	Blood pressure (after)	d=Before -After
1	169	162	7	-9.7778	95.6049
2	163	144	19	2.2222	4.9383
3	182	162	20	3.2222	10.3827
4	194	178	16	-0.7778	0.6049
5	172	146	26	9.2222	85.0494
6	193	176	17	0.2222	0.0494
7	147	140	7	-9.7778	95.6049
8	154	140	14	-2.7778	7.7160
9	175	150	25	8.2222	67.6049
			=151		367.5556
			= 151/9=16.7778

Sample mean of  paired differences :

standard deviation of the paired differences = 6.78

Step 3 of 5: Compute the value of the test statistic

value of the test statistic = 7.426

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H_o

Given = 0.01;

Degrees of freedom = n-1 =9-1=8

For Right Tailed Test : Reject null hypothesis if value of t-statistic is Greater than Critical Value ; t ₌ t_0.01

t_0.01 for 8 degrees of freedom = 2.896

Reject Ho if t > 2.896

Step 5 of 5: Make the decision for the hypothesis test.

As

t: 7.4258 > 2.896 reject H_o

Reject Null Hypothesis