Question

2. (5 pts) A medical researcher is testing a new drug to see if it slows the growth of malignant tumors in rats. The control group of 40 rats is given non of the drug and the test group of 40 rats is given a dose every day for two weeks. At the end of the test, the following statistics were calculated from the samples: Table 1: Control rats Average growth of tumor 6.2 grams Standard deviation 3.1 grams Test rats 4.1 grams 2.8 grams Given the assumption that the variances of the two population are equal, conduct the appropriate hypothesis test to determine if the new drug appears to slow the mean rate of tumor growth at a = 0.05.

Answer 1

Let the test rats by the population 1 and the control group by the population 2.

Now we have to test if the mean growth of the tumor is less in test rats than the control group rats.

hence we will conduct a t-test for comparing two means since the population standard deviation of the sample is not given
The provided sample means are shown below: X1 = 4.1 X2 = 6.2 Also, the provided sample standard deviations are: 81 = 2.8 $2 = 3.1 and the sample sizes are nı = 40 and n2 = 40. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: H1 H2 This corresponds to a left-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 = 78. 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 left-tailed test is te = -1.665, for a = 0.05 and df = 78. The rejection region for this left-tailed test is R = {t:t < -1.665}.

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X - X2 (nu - 1)3+(12-1)( + ) 4.1 - 6.2 = = -3.179 (40-1)2.82 +(40-1)3.1 40+402 ( 1 1 ) N 3.1 (10 + ) (4) Decision about the null hypothesis Since it is observed that t = -3.179 <te= -1.665, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0.0011, and since p= 0.0011 < 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 11 is less than 12, at the 0.05 significance level.

Since the null hypothesis is rejected we have evidence that the new drug slows the mean rate of tumor growth