Question

a marine biologist claims that the mean length of mature female pink seaperch is different in fall and 18 mature female pink seaperch collected in fall has a mean length of 111 millimeters and a standa millimeters. a sample of 8 mature female pink seaperch collected in winter has a mean length of 10 standard deviation of 9 millimeters. at α=0.20, can you support the marine biologist’s claim? assum variances are equal. assume the samples are random and independent, and the populations are no complete parts (a) through (e) below. 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 select the correct rejection region(s) below. a. t > t₀ b. t < -t₀, t > t₀ c. -t₀ < t < t₀ d. t < -t₀ (c) find the standardized test statistic. t = (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Identify given values

Let fall sample: \(n_1 = 18\), \(\bar{x}_1 = 111\), \(s_1\) (assuming typo, let's say \(s_1 = \) [missing? Wait, original problem might have \(s_1 = \) (let's check, maybe 10? Wait, original text: "18 mature... mean 111, standa [maybe 10?] and winter: \(n_2 = 8\), \(\bar{x}_2 = 10\) [no, 10? Wait, no, probably 105? Wait, typo in image. Wait, standard deviation for fall: let's assume \(s_1 = 10\) (since "standa" maybe 10), winter \(s_2 = 9\), \(\bar{x}_2 = 105\) (common typo). Wait, let's correct:

Assume:
Fall: \(n_1 = 18\), \(\bar{x}_1 = 111\), \(s_1 = 10\)
Winter: \(n_2 = 8\), \(\bar{x}_2 = 105\), \(s_2 = 9\)

Pooled variance: \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2 - 2}\)

Step2: Calculate pooled variance

\(s_p^2=\frac{(18 - 1)10^2+(8 - 1)9^2}{18 + 8 - 2}=\frac{17*100 + 7*81}{24}=\frac{1700 + 567}{24}=\frac{2267}{24}\approx94.4583\)
\(s_p=\sqrt{94.4583}\approx9.719\)

Step3: Calculate test statistic

\(t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
Since \(H_0: \mu_1 = \mu_2\), so \(\mu_1 - \mu_2 = 0\)
\(t=\frac{111 - 105}{9.719\sqrt{\frac{1}{18}+\frac{1}{8}}}=\frac{6}{9.719\sqrt{\frac{4 + 9}{72}}}=\frac{6}{9.719\sqrt{\frac{13}{72}}}\)
\(\sqrt{\frac{13}{72}}\approx\sqrt{0.1806}\approx0.425\)
\(9.719*0.425\approx4.131\)
\(t=\frac{6}{4.131}\approx1.452\) (Wait, maybe original \(s_1 = 12\)? Wait, let's recheck. Wait, maybe the fall standard deviation is 12? Let's try with \(s_1 = 12\):

\(s_p^2=\frac{17*144 + 7*81}{24}=\frac{2448 + 567}{24}=\frac{3015}{24}=125.625\), \(s_p=\sqrt{125.625}\approx11.21\)

\(t=\frac{111 - 105}{11.21\sqrt{\frac{1}{18}+\frac{1}{8}}}=\frac{6}{11.21\sqrt{0.1806}}\approx\frac{6}{11.21*0.425}\approx\frac{6}{4.764}\approx1.26\) No, maybe original \(\bar{x}_2 = 100\)? Wait, the image has "mean length of 10" which is typo. Let's assume correct values:

Wait, the problem says "mean length of 111" (fall), "mean length of 10" (winter) – no, that's impossible. So likely typo: winter mean is 105. Let's proceed with correct values (assuming \(\bar{x}_2 = 105\), \(s_1 = 12\), \(s_2 = 9\)):

Wait, maybe the original fall standard deviation is 12 (since "standa" could be 12). Let's recalculate:

\(s_p^2=\frac{(18-1)12^2 + (8-1)9^2}{24}=\frac{17*144 + 7*81}{24}=\frac{2448 + 567}{24}=\frac{3015}{24}=125.625\), \(s_p=\sqrt{125.625}=11.21\)

\(t=\frac{111 - 105}{11.21\sqrt{\frac{1}{18}+\frac{1}{8}}}=\frac{6}{11.21\sqrt{\frac{4 + 9}{72}}}=\frac{6}{11.21\sqrt{\frac{13}{72}}}\)

\(\sqrt{\frac{13}{72}}\approx0.425\), \(11.21*0.425\approx4.764\)

\(t=\frac{6}{4.764}\approx1.260\)

Wait, but maybe the fall standard deviation is 10. Let's try:

\(s_p^2=\frac{17*100 + 7*81}{24}=\frac{1700 + 567}{24}=2267/24≈94.458\), \(s_p≈9.719\)

\(t=\frac{111 - 105}{9.719\sqrt{\frac{1}{18}+\frac{1}{8}}}=\frac{6}{9.719*0.425}≈\frac{6}{4.131}≈1.452\)

But maybe the winter mean is 100. Let's check:

\(t=\frac{111 - 100}{s_p\sqrt{\frac{1}{18}+\frac{1}{8}}}\)

If \(s_1 = 12\), \(s_2 = 9\):

\(s_p^2=\frac{17*144 + 7*81}{24}=3015/24=125.625\), \(s_p=11.21\)

\(\sqrt{\frac{1}{18}+\frac{1}{8}}=\sqrt{\frac{4 + 9}{72}}=\sqrt{13/72}≈0.425\)

\(t=\frac{11}{11.21*0.425}≈\frac{11}{4.764}≈2.309\)

But the original problem's winter mean is likely 105 (since 10 is too low). Given the critical values are ±1.318, the test statistic should be calculated with correct values. Wait, maybe the fall sample standard deviation is 12, winter mean 105, winter standard deviation 9.

Wait, let's use the correct formula for two-sample t-test with equal variances:

\(t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\)

Assuming \(H_0: \mu_1 = \mu_2\), so \(\mu_1 - \mu_2 = 0\)

Let's take the values as:

Fall: \(n_1 = 18\), \(\bar{x}_1 = 111\), \(s_1 = 12\)

Winter: \(n_2 = 8\), \(\bar{x}_2 = 105\), \(s_2 = 9\)

Then:

\(s_p^2 = \frac{(18-1)(12)^2 + (8-1)(9)^2}{18 + 8 - 2} = \frac{17*144 + 7*81}{24} = \frac{2448 + 567}{24} = \frac{3015}{24} = 125.625\)

\(s_p = \sqrt{125.625} = 11.21\)

\(\sqrt{\frac{1}{18} + \frac{1}{8}} = \sqrt{\frac{4 + 9}{72}} = \sqrt{\frac{13}{72}} \approx 0.425\)

\(t = \frac{111 - 105}{11.21 * 0.425} = \frac{6}{4.764} \approx 1.260\)

But maybe the fall standard deviation is 10, winter mean 105:

\(s_p^2 = \frac{17*100 + 7*81}{24} = 2267/24 ≈ 94.458\)

\(s_p ≈ 9.719\)

\(t = \frac{6}{9.719 * 0.425} ≈ \frac{6}{4.131} ≈ 1.452\)

Alternatively, if the winter mean is 100:

\(t = \frac{11}{9.719 * 0.425} ≈ \frac{11}{4.131} ≈ 2.663\)

But given the critical value is 1.318, the test statistic should be calculated with the correct values. Wait, maybe the original problem has fall standard deviation 12, winter mean 105, winter standard deviation 9. Let's proceed with that, and the test statistic is approximately 1.26 (or 1.45, depending on s1). But maybe the correct values are:

Fall: \(n_1 = 18\), \(\bar{x}_1 = 111\), \(s_1 = 12\)

Winter: \(n_2 = 8\), \(\bar{x}_2 = 105\), \(s_2 = 9\)

Then the test statistic is approximately 1.26. But let's check with the pooled variance formula again.

Wait, maybe the user's original problem has fall standard deviation 12 (since "standa" could be 12), so:

\(s_p^2 = \frac{(18-1)(12)^2 + (8-1)(9)^2}{18+8-2} = \frac{17*144 + 7*81}{24} = \frac{2448 + 567}{24} = 3015/24 = 125.625\)

\(s_p = \sqrt{125.625} = 11.210\)

\(\sqrt{\frac{1}{18} + \frac{1}{8}} = \sqrt{\frac{4 + 9}{72}} = \sqrt{\frac{13}{72}} \approx 0.4247\)

\(t = \frac{111 - 105}{11.210 * 0.4247} = \frac{6}{4.761} \approx 1.260\)

So the standardized test statistic is approximately 1.260 (or 1.45, 2.31, etc., depending on correct values). But assuming the correct values (fall: 18, 111, 12; winter: 8, 105, 9), the test statistic is ~1.26.

Answer:

1.260 (or other correct value based on precise data; here, assuming the above calculation, the answer is approximately 1.260)