Question

a marine biologist claims that the mean length of mature female pink seaperch is different in fall and winter. a sample 18 mature female pink seaperch collected in fall has a mean length of 111 millimeters and a standard deviation of 15 millimeters. a sample of 8 mature female pink seaperch collected in winter has a mean length of 108 millimeters and a standard deviation of 9 millimeters. at α=0.20, can you support the marine biologist’s claim? assume the population variances are equal. assume the samples are random and independent, and the populations are normally distributed. complete parts (a) through (e) below. the alternative hypothesis, h_a the null hypothesis, h_0 (b) find the critical value(s) and identify the rejection region(s). enter the critical value(s) below. -1.318,1.318 (type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as neede select the correct rejection region(s) below. a. t > t_0 b. t < -t_0, t > t_0 c. -t_0 < t < t_0 d. t < -t_0

Explanation:

Step1: Identify Hypothesis Type

The alternative hypothesis \( H_a \) is that the means are different (\( \mu_1
eq \mu_2 \)), so it's a two - tailed test.

Step2: Determine Rejection Region

In a two - tailed t - test, the rejection regions are where the test statistic \( t \) is less than \( -t_0 \) or greater than \( t_0 \), where \( t_0 \) is the critical value. Given the critical values are - 1.318 and 1.318, the rejection region is \( t < - t_0 \) or \( t > t_0 \), which corresponds to option B.

Answer:

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