Question

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the 0.10 significance level, is there a difference in the population means?

(a) state the decision rule

(b) compute the pooled estimate of the population variance

(c) compute the test statistic

(d) state your decision about the null hypothesis

(e) estimate the p-value.

Answer 1

Answer:

a) $df =n_1 +n_2 -2 =15 +17-2 =30$

And then since we want to conduct a two tailed test we need to llok in the t distribution with 30 degrees of freedom a value who accumulate 0.05 of the area and we got:

$t_{\alpha/2} =\pm 1.697$

b)$S^2_p =\frac{(15-1)(12)^2 +(17 -1)(15)^2}{15 +17 -2}=187.2$

$S_p=13.682$

c) $t=\frac{(350 -342)-(0)}{13.682\sqrt{\frac{1}{15}+\frac{1}{17}}}=1.65$

d) For this case the statistic calculated is lower than the critical value so then we FAIL to reject the null hypothesis at the significance level given.

e) The p value for this case would be given by:

$p_v =2*P(t_{30}>1.65) =0.1094$

Step-by-step explanation:

We are assuming that the two population deviations are equal:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Part a

For this case we need to find the degrees of freedom given by:

$df =n_1 +n_2 -2 =15 +17-2 =30$

And then since we want to conduct a two tailed test we need to llok in the t distribution with 30 degrees of freedom a value who accumulate 0.05 of the area and we got:

$t_{\alpha/2} =\pm 1.697$

Part b

The pooled deviation can be founded with this formula:

$S^2_p =\frac{(15-1)(12)^2 +(17 -1)(15)^2}{15 +17 -2}=187.2$

And the deviation would be:

$S_p=13.682$

Part c

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Replacing the info given we got:

$t=\frac{(350 -342)-(0)}{13.682\sqrt{\frac{1}{15}+\frac{1}{17}}}=1.65$

Part d

For this case the statistic calculated is lower than the critical value so then we FAIL to reject the null hypothesis at the significance level given.

Part e

The p value for this case would be given by:

$p_v =2*P(t_{30}>1.65) =0.1094$