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 22 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)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) | 179 | 192 | 187 | 175 | 193 | 181 | 158 | 164 | 192 |
Blood pressure (after) | 171 | 179 | 177 | 163 | 183 | 164 | 149 | 148 |
186 |
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. Reject or Fail to Reject.
here, we will do paired t test
hypothesis:-
where , is the difference between blood pressure before taking new drug and blood pressure after taking new drug.
the necessary calculation table be:-
before() | after() | ||
179 | 171 | 8 | 10.38257 |
192 | 179 | 13 | 3.16057 |
187 | 177 | 10 | 1.49377 |
175 | 163 | 12 | 0.60497 |
193 | 183 | 10 | 1.49377 |
181 | 164 | 17 | 33.38297 |
158 | 149 | 9 | 4.93817 |
164 | 148 | 16 | 22.82737 |
192 | 186 | 6 | 27.27137 |
sum=101 | sum=105.55556 |
number of observations(n) = 9
the value of the standard deviation of the paired differences be:-
the test statistic be:-
[ for more accurate calculation i have used the sd = 3.6324 in spite of 6.3]
degrees of freedom = (n-1) = (9-1) = 8
t critical value for df = 8,alpha=0.01, right tailed test be:-
[from t distribution table]
decision rule:-
reject the null hypothesis if,
decision:-
so, we reject the null hypothesis.
conclusion:-
there is sufficient evidence to support the claim that its new drug reduces systolic blood pressure at 0.01 level of significance.
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