Question

Investigators want to determine the effectiveness of a new drug in lowering cholesterol. They conduct a small clinical trial in which patients were randomized to receive the new drug or placebo. The total cholesterol was measured after ten weeks on the assigned treatment. The table below shows the results.

  

Sample size (n) Mean Clostrol SD
New drug 30 198.8 26.7
Placebo 30   215.4 30.3

Is there sufficient evidence to conclude a difference in mean cholesterol for patients on the new drug compared to patients receiving the placebo? Use a significance level of ? = 0.05.

a) Give the name of the hypothesis test that would be appropriate for this situation. (1 point)

b) State the hypotheses in symbols. Make sure to define your variables first. (2 points)

c) Use your calculator to perform the appropriate hypothesis test and report the test statistic and p-value. Be sure to write out what you entered in your calculator. (3 points)

d) Write a full conclusion for this test in the context of the problem. (2 points)

Answer 1

The provided sample means are shown below: X1 = 198.8 12 215.4 Also, the provided sample standard deviations are: Si = 26.7 S2 = 30.3 and the sample sizes are ni 30 and n2 30. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: Mi 7 M2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 58. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is te = 2.002, for a = 0.05 and df = 58. The rejection region for this two-tailed test is R= {t : 1t| > 2.002}. : (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 t= (nı –1)s{+(n2-1)(+ m2) ni+n2-2 198.8 - 215.4 2.251 (30-1)26.72 +(30-1)30.32 30+30-2 1 30 + 30 (4) Decision about the null hypothesis Since it is observed that |t| = 2.251 >te = 2.002, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0282, and since p = 0.0282 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean Mi is different than 42, at the 0.05 significance level.

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