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 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.

c. “the mean length of mature female pink seaperch is different in fall and winter.”
d. “the mean length of mature female pink seaperch is greater in the winter than in the fall.”
what are h₀ and hₐ?
the null hypothesis, h₀, is μ₁ = μ₂. the alternative hypothesis, hₐ, is μ₁ ≠ μ₂.
which hypothesis is the claim?
the alternative hypothesis, hₐ
the null hypothesis, h₀
(b) find the critical value(s) and identify the rejection region(s).
enter the critical value(s) below.

Explanation:

Step1: Identify Test Type

This is a two - sample t - test (since population variances are equal, samples are independent, random, and populations are normal) with a two - tailed test (because the claim is that the means are different, \(H_a:\mu_1
eq\mu_2\)). The significance level \(\alpha = 0.20\). The degrees of freedom \(df=n_1 + n_2-2=18 + 8-2 = 24\).

Step2: Find Critical Values

For a two - tailed t - test with \(\alpha = 0.20\) and \(df = 24\), we can use the t - distribution table or a calculator. The critical values are \(t_{\alpha/2,df}\) and \(-t_{\alpha/2,df}\). \(\alpha/2=0.10\), so we look for \(t_{0.10,24}\). From the t - table or using a calculator (e.g., \(t=\text{T.INV.2T}(0.20,24)\)), we get \(t=\pm1.318\).

Answer:

\(- 1.318,1.318\)