Answer to Question #290938 in Statistics and Probability for Shankar

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

"n_a=5\\\\\\bar x_a={\\sum x\\over n_a}={230\\over5}=46"

"s_a^2={\\sum x^2-{(\\sum x)^2\\over n_a}\\over n_a-1}={10870-10580\\over4}=72.5"

Medicine type B

"n_b=5\\\\\\bar x_b={\\sum x\\over n_b}={235\\over5}=47"

"s_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,

"H_0:\\sigma_1^2=\\sigma_2^2\\\\vs\\\\H_1:\\sigma_1^2\\not=\\sigma_2^2"

The test statistic is,

"F_c={s_b^2\\over s_a^2}={139\\over72.5}=1.91724138"

The table value is,

"F_{\\alpha,n_b-1,n_a-1}=F_{0.05,4,4}=6.388233" and we reject the null hypothesis if "F_c\\gt F_{\\alpha,n_b-1,n_a-1}"

Since "F_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,

"H_0:\\mu_a=\\mu_b\\\\vs\\\\H_1:\\mu_a\\not=\\mu_b"  

The test statistic is,

"t_c={(\\bar x_a-\\bar x_b)\\over \\sqrt{sp^2({1\\over n_a}+{1\\over n_b})}}"

where "sp^2" is the pooled sample variance given as,

"sp^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,

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

"t_c" is compared with the table value at "\\alpha=0.05" with "n_a+n_b-2=5+5-2=8" degrees of freedom.

The table value is,

"t_{{0.05\\over2},8}=t_{0.025,8}=2.306004"

The null hypothesis is rejected if "|t_c|\\gt t_{0.025,8}."

Now,

"|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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