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). a α=0.10, 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 | 392 445 397 449 389 440 376
conventional | 354 357 385 354 364 394 404 391 438 443

(b) find the critical value(s) and identify the rejection region(s).
enter the critical value(s) below.
-1.753,1.753
(type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as need
select the correct rejection region(s) below.
a. t < -t₀
b. t > t₀
c. -t₀ < t < t₀
d. t < -t₀, t > t₀

Explanation:

Step1: Identify Test Type

This is a two - sample t - test (population variances equal) to compare two treatments. The alternative hypothesis is two - tailed (we want to know if there is a difference, not a specific direction), so the rejection region will be in both tails.

Step2: Analyze Rejection Regions

• For a two - tailed test with significance level \(\alpha = 0.10\), the critical values are \(\pm t_{\alpha/2}\). The rejection region is where the test statistic \(t\) is less than \(-t_{0}\) or greater than \(t_{0}\), which corresponds to option D.

Answer:

D. \(t < - t_{0},t>t_{0}\)