Question #290938

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]

1
Expert's answer
2022-01-30T13:20:03-0500

Medicine type A

na=5xˉa=xna=2305=46n_a=5\\\bar x_a={\sum x\over n_a}={230\over5}=46

sa2=x2(x)2nana1=10870105804=72.5s_a^2={\sum x^2-{(\sum x)^2\over n_a}\over n_a-1}={10870-10580\over4}=72.5

Medicine type B

nb=5xˉb=xnb=2355=47n_b=5\\\bar x_b={\sum x\over n_b}={235\over5}=47

sb2=x2(x)2nbnb1=11601110454=139s_b^2={\sum x^2-{(\sum x)^2\over n_b}\over n_b-1}={11601-11045\over4}=139

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

We test,

H0:σ12=σ22vsH1:σ12σ22H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2

The test statistic is,

Fc=sb2sa2=13972.5=1.91724138F_c={s_b^2\over s_a^2}={139\over72.5}=1.91724138

The table value is,

Fα,nb1,na1=F0.05,4,4=6.388233F_{\alpha,n_b-1,n_a-1}=F_{0.05,4,4}=6.388233 and we reject the null hypothesis if Fc>Fα,nb1,na1F_c\gt F_{\alpha,n_b-1,n_a-1}

Since Fc=1.91724138<F0.05,4,4=6.388233F_c=1.91724138\lt F_{0.05,4,4}=6.388233, we accept the null hypothesis that the population variances for medicine type A and medicine type B are equal.


We now proceed to test whether the difference in means is significant.

The hypothesis tested are,

H0:μa=μbvsH1:μaμbH_0:\mu_a=\mu_b\\vs\\H_1:\mu_a\not=\mu_b  

The test statistic is,

tc=(xˉaxˉb)sp2(1na+1nb)t_c={(\bar x_a-\bar x_b)\over \sqrt{sp^2({1\over n_a}+{1\over n_b})}}

where sp2sp^2 is the pooled sample variance given as,

sp2=(na1)sa2+(nb1)sb2na+nb2=(4×72.5)+(4×139)8=8468=105.75sp^2={(n_a-1)s_a^2+(n_b-1)s_b^2\over n_a+n_b-2}={(4\times72.5)+(4\times139)\over8}={846\over8}=105.75

Therefore,

tc=(4647)105.75(15+15)=16.5038=0.1538t_c={(46-47)\over \sqrt{105.75({1\over 5}+{1\over 5})}}={-1\over6.5038}=-0.1538

tct_c is compared with the table value at α=0.05\alpha=0.05 with na+nb2=5+52=8n_a+n_b-2=5+5-2=8 degrees of freedom.

The table value is,

t0.052,8=t0.025,8=2.306004t_{{0.05\over2},8}=t_{0.025,8}=2.306004

The null hypothesis is rejected if tc>t0.025,8.|t_c|\gt t_{0.025,8}.

Now,

tc=0.1538<t0.025,8=2.306004,|t_c|=0.1538\lt t_{0.025,8}=2.306004, and we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the  two medicines differ significantly with regard to their effect and increasing weight at 5% significance level.


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