Answer to Question #257992 in Statistics and Probability for hari

Question

In a certain experiment to compare two types of drugs A and B, the following results of increase in weights (in kg.) were observed in 5 persons. Person number 1 2 3 4 5 Increase weight in kg. Drug A 3.6 5.5 5.9 4.1 1.4 Drug B 4.5 3.6 5.5. 6.8 2.7 (i) Assuming that the two samples of persons are independent; can we conclude that the drug B is better than drug A? (ii) Also, examine the case when the same set of 5 persons was used in both the drugs.

Answer 1

Using TI84, the provided sample means are shown below:

$\bar X_1 = 4.1$ $\bar X_2$ =4.529

Also, the provided sample standard deviations are :$s _1 ​ =1.785; s_2 = 1.378$

and the sample sizes are $n_1 = 5 and$ $n_2 = 7$

The following null and alternative hypotheses need to be tested:

$Ho: \mu_1= \mu_2$

$Ha: \mu_1≠ \mu_2$

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.

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = $df=\frac{\frac{s_{1}^{2}}{n_1}+\frac{s_{2}^{2}}{n_2}}{\frac{\frac{s_{1}^{2}}{n_1}}{n_1-1}+\frac{\frac{s_{2}^{2}}{n_2}}{n_2-1}}$ =$df=\frac{\frac{1.785^{2}}{5}+\frac{1.378^{2}}7}{\frac{\frac{1.785^{2}}5}{5-1}+\frac{\frac{1.378^{2}}{7}}{7-1}}$= 7.254

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

$t = \frac{\overline{X1}-\overline{X2}}{\sqrt{\frac{(s1^2)}{n1}+\frac{s2^2}{n2}}}$ = $\frac{4.1-4..529}{\sqrt{\frac{(1.7851^2)}{5}+\frac{1.378^2}{7}}}$ = -0.45

Hence, it is found that the critical value for this two-tailed test is tc

​=2.348, for α=0.05 and df = 7.254, using t table the rejection region for this two-tailed test is R={t:∣t∣>2.348}.since critical value is greater than test statistic ,it is concluded that the null hypothesis is not rejected.

t is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is different than μ2​, at the 0.05 significance level.