Consider the following hypothesis test:

$H_0:μ_1-μ_2=0$

$H_a:μ_1-μ_2≠0$

The following results are from independent sample taken from two populations:

What is the rejection rule using the critical value? What is your conclusion? ($α = 0.05$)

Reject $H_0$ if $∣t∣>1.997$; Since $∣2.18∣>1.997$, reject $H_0$

The correct answer is Option (1) → Reject $H_0$ if $∣t∣>1.997$; Since $∣2.18∣>1.997$, reject $H_0$

Given, $D_0 = 0, n_1 = 35, n_2 = 40, x_1 = 13.6, x_2 = 10.1, S_1 = 5.2,S_2 = 8.5$ and $α = 0.05$

$t = \frac{(\bar{x}_1 - \bar{x}_2)-D_0}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} = \frac{(13.6 - 10.1)-0}{\sqrt{\frac{(5.2)^2}{35} + \frac{(8.5)^2}{40}}}$

$= \frac{3.5}{\sqrt{\frac{27.04}{35} + \frac{72.25}{40}}} = \frac{3.5}{\sqrt{0.77251 + 1.80625}} = \frac{3.5}{\sqrt{2.57876}}$

$= \frac{3.5}{1.6058}=2.18$

$∴t=2.18$

$df=\frac{\left( \frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} \right)^2}{\frac{1}{n_1 - 1}\left( \frac{S_1^2}{n_1} \right)^2 + \frac{1}{n_2 - 1}\left( \frac{S_2^2}{n_2} \right)^2}$

$=\frac{\left(\frac{(5.2)^2}{35} + \frac{(8.5)^2}{40}\right)}{\frac{1}{35-1}\left(\frac{(5.2)^2}{35}\right)^2+\frac{1}{40-1}\left(\frac{(8.5)^2}{40}\right)^2}$

$=\frac{(0.77251+1.80625)^2}{\frac{1}{34}(0.77251)^2+\frac{1}{39}(1.80625)^2}=\frac{6.6503}{\frac{1}{34}(0.5969)+\frac{1}{39}(3.2625)}$

$=\frac{6.6503}{0.0176+0.0837}=\frac{6.6503}{0.1013}=65.64$

So $df=65$

Reject $H_0$ if $t≤-t_{α/2}$ or $t≥t_{α/2}$

Here, $t = 2.18$ and $t_{α/2} = 0.025$

From the table, $t_{0.025} = 1.997$ with $df = 65$.

$∵ 2.18 > 1.997$

So, reject $H_0$.