Answer to Question #340822 in Statistics and Probability for Christopher

Question

The diameter of steel rods manufactured on two different extrusion machines are

being investigated. Two random samples of sizes $n_1$ = 15 and $n_2$ = 17 are selected,

and the sample means and sample variances are x̅1 = 8.73, $s_1^2$ = 0.35, x̅2 =8.68, and $s_2^2$ = 0.40, respectively. Assume that $\sigma_1^2=\sigma_2^2$

and that data are drawn from

a normal distribution. (a) Is there evidence to support the claim that the two

machines produce rods with different mean diameters? Use α = 0.05 in arriving at

this conclusion. (b) Find the P-value for the t-statistic you calculated in part (a).

[taken from Montgomery, p. 347]

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=df_1+df_2=n_1+n_2-2=30$$

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df = 30.$

The critical value for this two-tailed test, $\alpha = 0.05, df=30$ degrees of freedom is $t_c =2.042272.$

The rejection region for this two-tailed test is $R = \{t: |t|> 2.042272\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{8.73-8.68}{\sqrt{\dfrac{(15-1)(0.35)+(17-1)(0.40)}{15+17-2}(\dfrac{1}{15}+\dfrac{1}{17})}}$$

$$=0.229978$$

Since it is observed that $|t|=0.229978<2.042272= t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=30$ degrees of freedom, $t=0.229978$ is $p= 0.81967,$ and since $p=0.81967>0.05=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.05$ significance leve