Question

12. Test the claim that ul = u2. Two samples are randomly selected and come from populations that are normal. The sample statistics are given below. Assume that o2 equal o"2 (2). Use a = 0.05. nl=25 xbarl-30 .s1= 1.5 n2-30 xbar2-28 s2=1.9

Answer 1

Null Hypothesis : $H_0:\mu_1=\mu_2$

Alternative Hypothesis : $H_1:\mu_1\neq \mu_2$

We will test this hypothesis at 0.05 significance level.

The standard error would be given by,

$SE=\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_1^2}{n_2} \right )}=\sqrt{\left(\frac{1.5^2}{25}+\frac{1.9^2}{30} \right )}=0.46$

The test statistic would be given by,

$t=\frac{\bar{x}_1-\bar{x}_2}{SE}=\frac{30-28}{}{0.46}=4.36$

The degree of freedom would be given by,

$df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_1^2}{n_2} \right )^2}{\frac{1}{n_1}\left (\frac{s_1^2}{n_1} \right )^2+\frac{1}{n_2}\left (\frac{s_2^2}{n_2} \right )^2}=\frac{\left(\frac{1.5^2}{25}+\frac{1.9^2}{30} \right )^2}{\frac{1}{24}\left (\frac{1.5^2}{25} \right )^2+\frac{1}{29}\left (\frac{1.9^2}{30} \right )^2}\approx 53$

The critical value will be the value of t-distribution with 53 degrees of freedom and 0.975 cumulative probability.

$t_{cr}=2.01$

Since the test statistic is greater than critical value we will reject the null hypothesis. Thus we do not have enough evidence to support the claim that the mean of the two groups are equal.