Answer to Question #290934 in Statistics and Probability for Shankar

Question

A group of 5 patients treated with Medicine type A weight 42, 39, 48, 60 and 41 kg.

A second group of 5 patients treated with Medicine type B weight 38, 42, 48, 67, 40 kg. Do

the two medicines differ significantly with regard to their effect and increasing weight? [At

5% level of significance]

Answer 1

$$\bar{x}_1=\dfrac{42+39+48+60+41}{5}=46$$ $$s_1^2=\dfrac{1}{5-1}((42-46)^2+(39-46)^2+(48-46)^2$$ $$+(60-46)^2+(41-46)^2)=\dfrac{145}{2}$$

$$s_1=\sqrt{s_1^2}=\sqrt{\dfrac{145}{2}}\approx8.5147$$

$$\bar{x}_2=\dfrac{38+42+48+67+40}{5}=47$$ $$s_2^2=\dfrac{1}{5-1}((38-47)^2+(42-47)^2+(48-47)^2$$ $$+(67-47)^2+(40-47)^2)=139$$

$$s_2=\sqrt{s_2^2}=\sqrt{139}\approx11.7898$$

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{72.5}{139}=0.5216$$

The critical values for two-tailed, $df_1=df_2=5-1=4$ are $F_L = 0.1041$  and $F_U = 9.6045,$ and since $F = 0.5216,$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

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.

The significance level is $\alpha = 0.05,$ and the degrees of freedom are $df =n_1+n_2-2= 5+5-2=8$

Hence, it is found that the critical value for this two-tailed test, $\alpha = 0.05$ and $df = 8$ is $t_c = 2.3060.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.3060\}.$

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

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{46-47}{\sqrt{\dfrac{(5-1)(72.5)+(5-1)(139)}{5+5-2}(\dfrac{1}{5}+\dfrac{1}{5})}}$$

$$=-0.153755$$

Since it is observed that $|t| = 0.153755 \le 2.3060=t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=8$ degrees of freedom, $t=-0.153755,$ is $p=0.881611,$ and since $p = 0.881611 \ge 0.05=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.