Question

A group of 5 patients treated with medicine A weigh 42, 39, 48, 60, 41 kg. A second group of 7 patients from the same hospital treated with medicine B weigh 38, 42, 56, 64, 68, 69, 62kg. Do you agree with the claim that the medicine B increases the weight significantly t= 2.2281

Answer 1

Answer:

please have a look in the attached images.

Answer 2

Answer:

Conclude that the medicines $\mathrm{A} and also \mathrm{B}$ do not differ significantly.

Explanation:

Given: A group of 5 patients treated with medicine A weigh 42, 39, 48, 60, 41 kg. A second group of 7 patients from the same hospital treated with medicine B weigh 38, 42, 56, 64, 68, 69, 62kg.

To find: Claim the medicine increases the weight

Ho:., $\mu_{1}=\mu_{2}$ (i.e) there is no significant difference between the medicines A and B as regards on increase in weight.

$\mathrm{H}_{1} \mu_{1 \neq} \mu_{2}$ (i.e) there is a significant difference between the medicines $\mathrm{A} and \mathrm{B}$

Level of significance $=5 \%$

Before we go to test the means first we have to test their variability using F-test.

F-test

Ho: $., \sigma_{1}^{2}=\sigma_{2}^{2}$

$\mathrm{H} 1:, \sigma_{1}^{2} \neq \sigma_{2}^{2}$

$S_{1}^{2}=\frac{\sum x_{1}^{2}-\frac{\left(\sum x_{1}\right)^{2}}{n 1}}{n 1-1}=82.5 \\ S_{2}^{2}=\frac{\sum x_{2}^{2}-\frac{\left(\sum x_{2}\right)^{2}}{n 2}}{n 2-1}=154.33$

$\therefore F=\frac{S_{2}^{2}}{S_{1}^{2}} \sim F_{\left(n_{2}-1, n_{1-} 1\right)} \text { d.f if } S_{2}^{2} > S_{1}^{2}$

$\begin{gathered}F_{\text {cal }}=\frac{154.33}{32.5}=1.8707 \\\mathrm{~F}_{\mathrm{tab}}(6,4) \mathrm{d} . \mathrm{f}=6.16 \\\Rightarrow \mathrm{F}_{\mathrm{cal}} < \mathrm{F}_{\mathrm{tab}}\end{gathered}$

We accept the null hypothesis $\mathrm{H}_{0}$.(i.e) the variances are equal.

Test statistic

$t=\frac{{\left|\left(\bar{x}_{1}-\bar{x}_{2}\right)\right|}}{S \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\\\\\sim t_{(n 1+n 2-2)} d . f$

Where

$S^{2}=\frac{\left[\sum x_{1}^{2}-\frac{\left(\sum x_{1}\right)^{2}}{n 1}\right]+\left[\sum x_{2}^{2}-\frac{\left(\sum x_{2}\right)^{2}}{n 2}\right]}{n_{1}+n_{2}-2}\\=\frac{330+926}{10}\\=125.6$

$t=\frac{|44-57|}{\sqrt{125.6\left(\frac{1}{7}+\frac{1}{75}\right)}}=1.98$

Table value

$\mathrm{t}_{\mathrm{tab}[(5+7-2)=10]} \mathrm{d} .f at 5 \% l.o.s =2.228$

Inference:

$\mathrm{t}_{\mathrm{cal}} < \mathrm{t}_{\mathrm{tab}}$

We accept the null hypothesis $\mathrm{H}_{0}$

We conclude that the medicines $\mathrm{A} and also \mathrm{B}$ do not differ significantly.