Question

A test among three groups of students shows significance with respect to differences in their GPAS. The mean of group 1 is 3.
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Answer #1

Degrees of freedom = n-k = 100 - 3 = 97

Where

n = 40+25+35 = 100

K = No. of groups = 3

Hence critical value = 1.985

It is obtained using R software, with command qt(0.975,97)

1. Group 1 vs group 2

Test statistic,

\small \\ t = \frac{\bar{X}_1-\bar{X}_2}{\sqrt{S^2\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}\\\\\\ t = \frac{3.1-2.9}{\sqrt{0.7*\left ( \frac{1}{40}+\frac{1}{25} \right )}} \\\\\\ t = 0.94

Fail to reject null hypothesis that \small \mu_1 =\mu_2 and conclude that \small \mu_1 \neq \mu_2

2. Group 1 vs group 3

Test statistic,

\small \\ t = \frac{\bar{X}_1-\bar{X}_3}{\sqrt{S^2\left ( \frac{1}{n_1}+\frac{1}{n_3} \right )}}\\\\\\ t = \frac{3.1-3}{\sqrt{0.7*\left ( \frac{1}{40}+\frac{1}{35} \right )}} \\\\\\ t = 0.52

Fail to reject null hypothesis that \small \mu_1 =\mu_3 and conclude that \small \mu_1 \neq \mu_3 ​​​​​​​​​​​​​​

3. Group 2 vs group 3

Test statistic,

\small \\ t = \frac{\bar{X}_2-\bar{X}_3}{\sqrt{S^2\left ( \frac{1}{n_2}+\frac{1}{n_3} \right )}}\\\\\\ t = \frac{2.9-3}{\sqrt{0.7*\left ( \frac{1}{25}+\frac{1}{35} \right )}} \\\\\\ t = -0.46

Positive test statistic, t =0.46

Fail to reject null hypothesis that \small \mu_2 =\mu_3 and conclude that \small \mu_2 \neq \mu_3 ​​​​​​​​​​​​​​​​​​​​​

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