Question

a marine biologist claims that the mean length of mature female pink seaperch is different in fall and winter. a sample of 15 mature female pink seaperch collected in fall has a mean length of 108 millimeters and a standard deviation of 9 millimeters. a sample of 8 mature female pink seaperch collected in winter has a mean length of 103 millimeters and a standard deviation of 8 millimeters. at \\( \alpha = 0.02 \\), can you support the marine biologists 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. use a comma to separate answers as needed.) select the correct rejection region(s) below. a. \\( t < - t _ { 0 } \\) b. \\( t < - t _ { 0 }, t > t _ { 0 } \\) c. \\( t > t _ { 0 } \\) d. \\( - t _ { 0 } < t < t _ { 0 } \\) (c) find the standardized test statistic. \\( t = 1.316 \\) (type an integer or decimal rounded to three decimal places as needed.) (d) decide whether to reject or fail to reject the null hypothesis. fail to reject the null hypothesis. (e) interpret the decision in the context of the original claim. at the 2% significance level, enough evidence to support the claim that the mean lengths of mature female pink seaperch are different in fall and winter.

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n_1 + n_2-2\), where \(n_1 = 15\) and \(n_2=8\). So \(df=15 + 8-2=21\).

Step2: Find the critical value \(t_0\)

Since \(\alpha = 0.02\) and it is a two - tailed test, we look up the \(t\) - value in the \(t\) - distribution table. For \(\alpha/2=0.01\) and \(df = 21\), \(t_0\approx 2.518\).

Step3: Analyze the test statistic

The test statistic \(t = 1.316\). Since \(|t|=1.316

Answer:

At the \(2\%\) significance level, there is not enough evidence to support the claim that the mean lengths of mature female pink seaperch are different in fall and winter.