Solution to A group of 5 patients treated with Medicine type A weight 42, 39, 48, 60 … - Sikademy
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Peace Awoke

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]

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


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:



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:




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.

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