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 ?=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) 182 164 176 171 197 168 176 177 187 Blood pressure (after) 175 151 153 159 179 143 160 161 162 Step 1 of 5 : State the null and alternative hypotheses for the test.

Answer 1

Differences (Blood pressure (before) - Blood pressure (after)): 7, 13, 23, 12, 18, 25, 16, 16, 25 We calculate sample mean and std deviation from given data. Sample Mean, d^bar = sigma(d)/n = 155/9 = 17.222222 Sample Variance, s^2 = sigma(d - d^bar)^2/n - 1 = 307.555556/8 = 38.444444 Sample std dev, s = squareroot s^2 = squareroot 38.444444 almostequalto 6.200358 Test: mu_D > 0 with t dist^n Hypothesis test: H_0: mu_D = 0 H_a: mu_D > 0 (Null Hypothesis, H_0: mu_D lessthanorequalto 0 is also correct) (Alternative Hypothesis, also called H_1) This is upper tailed test. d^bar = 17.222222 s = 6.200358 n = 9 Significance Level, alpha = 0.05 Degrees of freedom df = n - 1 = 8 t_observed = d^bar - mu_D/S/squareroot n = 17.222222 - 0/6.200358/squareroot 9 almostequalto 8.333 Test Statistic, t_0 = 8.333 Using t-table, we can observe that the p-value is less than 0.0005 \r\n Using excel we find exact p-value = P(t > 8.333) = t.dist(-8.333, 8, 1) = 0.0000162627