In: Statistics and Probability
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 = 0.05 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)
| 199 | 166 | 183 | 197 | 200 | 192 | 190 | 179 | 200 |
Blood pressure (after)
| 183 | 151 | 172 | 174 | 185 | 170 | 180 | 173 | 185 |
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 H0. Round the numerical portion of your answer to three decimal places.
Step 5 of 5: Make the decision for the hypothesis test.
| Before | After | Difference |
| 199 | 183 | 16 |
| 166 | 151 | 15 |
| 183 | 172 | 11 |
| 197 | 174 | 23 |
| 200 | 185 | 15 |
| 192 | 170 | 22 |
| 190 | 180 | 10 |
| 179 | 173 | 6 |
| 200 | 185 | 15 |
| Total | 133 |
Step 1:
A pharmaceutical company claims that its new drug reduces systolic blood pressure.
The null and alternative hypothesis is


Step 2:
Sample standard deviation of difference =
= 5.4
Step 3:
Sample size = n = 9
Sample mean of difference =
= 14.78
Test statistic is


Step 4:
Degrees of freedom = n - 1 = 9 - 1 = 8
Critical value = 1.860 ( Using t table)
Step 5:
Test statistic > critical value we reject null hypothesis.
Conclusion:
A pharmaceutical company claims that its new drug reduces systolic blood pressure is right.