Final answer:
The correct answer is D. The researcher suspects that the new drug results in greater average reduction in blood pressure than the old drug. The P-value for the test described in (again, using Option 2 for degrees of freedom) is between 0.02 and 0.025.
Step-by-step explanation:
To test the hypothesis that the new drug results in a greater average reduction in blood pressure than the old drug, we can perform a two-sample t-test. The null hypothesis is that there is no difference in the average reduction between the two drugs, while the alternative hypothesis is that the new drug has a higher average reduction.
The test statistic is calculated as:
T = (xbar1 - xbar2) / sqrt((s1^2/n1) + (s2^2/n2))
where xbar1 and xbar2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. Using the given data:
T = (23.48 - 18.52) / sqrt((8.01^2/21) + (7.15^2/23)) ≈ 2.03
The degrees of freedom for this test is the smaller sample size minus 1, so df = 21 - 1 = 20. Using a t-distribution table or statistical software, we can find the p-value associated with this test statistic and degrees of freedom. Based on the information given, the P-value for the test is between 0.02 and 0.025, therefore the correct answer is D.