Question

an engineer wants to know if producing metal bars using a new experimental treatment rather than the conventional treatment makes a difference in the tensile strength of the bars (the ability to resist tearing when pulled lengthwise). at $\alpha = 0.02$, answer parts (a) through (e). assume the population variances are equal and the samples are random. if convenient, use technology to solve the problem.
treatment tensile strengths (newtons per square millimeter)
experimental 365 353 432 363 385 413 386
conventional 444 394 411 374 389 362 350 416 434 447
the alternative hypothesis, $h_a$
(b) find the critical value(s) and identify the rejection region(s).
enter the critical value(s) below.
$\pm 2.602$
(type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as needed.)
select the correct rejection region(s) below.
$\bigcirc$ a. $-t_0 < t < t_0$
$\bigcirc$ b. $t < -t_0$
$\bigcirc$ c. $t > t_0$
$\bigcirc$ d. $t < -t_0, t > t_0$

Explanation:

Step1: Determine the type of test

Since we are testing if there is a difference (not specifying greater or less), it is a two - tailed test.

Step2: Recall the properties of the t - distribution for two - tailed tests

For a two - tailed t - test with significance level \(\alpha = 0.02\) and degrees of freedom \(df=n_1 + n_2-2\). Here \(n_1 = 7\) (experimental group) and \(n_2=10\) (conventional group), so \(df=7 + 10-2=15\). The critical values for a two - tailed t - test with \(\alpha = 0.02\) and \(df = 15\) are \(t=\pm t_{\alpha/2,df}\). Using a t - table or technology, \(t_{\alpha/2,df}=t_{0.01,15}\approx\pm 2.602\).

Step3: Identify the rejection regions for a two - tailed test

In a two - tailed t - test, the rejection regions are \(t < - t_0\) and \(t>t_0\)

Answer:

D. \(t < - t_0,t > t_0\)