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. the null hypothesis, \\( h _ { 0 } \\) the alternative hypothesis, \\( h _ { a } \\) (b) find the critical value(s) and identify the rejection region(s). enter the critical value(s) below. \\( \pm 2.518 \\) (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. 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 = \square \\) (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Calculate the pooled variance

The formula for pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Here, \(n_1 = 15\), \(s_1=9\), \(n_2 = 8\), \(s_2 = 8\)
\(s_p^2=\frac{(15 - 1)\times9^2+(8 - 1)\times8^2}{15 + 8-2}=\frac{14\times81+7\times64}{21}=\frac{1134 + 448}{21}=\frac{1582}{21}\approx75.333\)

Step2: Calculate the standardized test statistic \(t\)

The formula for \(t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}}\)
\(\bar{x}_1 = 108\), \(\bar{x}_2=103\)
\(t=\frac{108 - 103}{\sqrt{75.333(\frac{1}{15}+\frac{1}{8})}}=\frac{5}{\sqrt{75.333\times(\frac{8 + 15}{120})}}=\frac{5}{\sqrt{75.333\times\frac{23}{120}}}\)
First, calculate \(75.333\times\frac{23}{120}\approx14.406\)
Then \(\sqrt{14.406}\approx3.796\)
\(t=\frac{5}{3.796}\approx1.317\)

Answer:

\(t = 1.317\)