Question

a researcher claims that the stomachs of blue crabs from location a contai from location b. the stomach contents of a sample of 13 blue crabs from l of fish and a standard deviation of 35 milligrams. the stomach contents of a contain a mean of 181 milligrams of fish and a standard deviation of 41 mill the researcher’s claim? assume the population variances are equal. compl (a) identify the null and alternative hypotheses. choose the correct answer \\(\bigcirc\\) a. \\(h_0: \mu_1 - \mu_2 \geq 0\\) \\(h_a: \mu_1 - \mu_2 < 0\\) \\(\bigcirc\\) c. \\(h_0: \mu_1 - \mu_2 < 0\\) \\(h_a: \mu_1 - \mu_2 = 0\\) \\(\bigcirc\\) b. \\(h_0: \mu_1\\) \\(h_a: \mu_1\\) \\(\bigcirc\\) d. \\(h_0: \mu_1\\) \\(h_a: \mu_1\\) (b) find the standardized test statistic for \\(\mu_1 - \mu_2\\). \\(t = \square\\) (round to three decimal places as needed.)

Explanation:

Step1: Identify Missing Values

Assume the sample from Location A has \( n_1 = 13 \), \( \bar{x}_1 \) (missing, but likely from context, let's assume the other sample: suppose Location B has \( n_2 \), \( \bar{x}_2 = 181 \), \( s_2 = 41 \). Wait, the original text is cut, but typical two - sample t - test with equal variances: formula for test statistic \( t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)_0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \), where \( s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}} \), and \( (\mu_1 - \mu_2)_0 = 0 \) (from null hypothesis of equality or difference). Let's assume the missing values: from the visible part, maybe Location A: \( n_1 = 13 \), \( \bar{x}_1 \) (let's say the other sample, maybe the first sample has \( \bar{x}_1 \) (maybe the problem had a typo, but let's assume the common case where we have two samples. Wait, the original problem's first part: "the stomachs of blue crabs from Location A contain... from Location B. The stomach contents of a sample of 13 blue crabs from [Location A?] of fish and a standard deviation of 35 milligrams. The stomach contents of [Location B?] contain a mean of 181 milligrams of fish and a standard deviation of 41 milligrams." Let's assume \( n_1 = 13 \), \( \bar{x}_1 \) (let's say the mean for A is, maybe the problem had a missing number, but perhaps in the original problem, the mean for A is, for example, if we assume the claim is that A has less than B, but let's suppose the correct values: Let's assume \( n_1 = 13 \), \( \bar{x}_1 = 150 \) (hypothetical, but since the problem is cut, but let's proceed with the formula. Wait, maybe the original problem has \( n_1 = 13 \), \( \bar{x}_1 \), \( s_1 = 35 \); \( n_2 \) (let's say \( n_2 = 14 \) (common sample size), \( \bar{x}_2 = 181 \), \( s_2 = 41 \). But since the problem is incomplete, but let's assume the correct approach.

Step2: Calculate Pooled Standard Deviation

First, we need \( n_1 \), \( s_1 \), \( n_2 \), \( s_2 \). Let's assume \( n_1 = 13 \), \( s_1 = 35 \); \( n_2 = 14 \), \( \bar{x}_2 = 181 \), \( s_2 = 41 \), and \( \bar{x}_1 \) (let's say the mean for A is, for example, 150 (just to demonstrate, but the actual value is missing). Wait, this is a problem with the input. But since the user provided a partial problem, but let's assume that in the original problem, the mean for Location A is, say, 150 (this is a guess, but to show the method).

Let \( n_1 = 13 \), \( \bar{x}_1 = 150 \), \( s_1 = 35 \); \( n_2 = 14 \), \( \bar{x}_2 = 181 \), \( s_2 = 41 \).

First, calculate \( s_p \):

\( (n_1 - 1)s_1^2=(13 - 1)\times35^2=12\times1225 = 14700 \)

\( (n_2 - 1)s_2^2=(14 - 1)\times41^2=13\times1681 = 21853 \)

\( n_1 + n_2-2=13 + 14 - 2=25 \)

\( s_p=\sqrt{\frac{14700 + 21853}{25}}=\sqrt{\frac{36553}{25}}=\sqrt{1462.12}\approx38.238 \)

Then, the standard error \( SE = s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=38.238\sqrt{\frac{1}{13}+\frac{1}{14}}=38.238\sqrt{\frac{14 + 13}{13\times14}}=38.238\sqrt{\frac{27}{182}}\approx38.238\times0.387\approx14.80 \)

The test statistic \( t=\frac{(\bar{x}_1-\bar{x}_2)-0}{SE}=\frac{150 - 181}{14.80}=\frac{- 31}{14.80}\approx - 2.095 \)

But this is with assumed values. Since the problem's text is cut, we need the correct sample means and sample sizes. However, if we assume the correct values (for example, if the mean for Location A is, say, 160, \( n_1 = 13 \), \( n_2 = 14 \)):

\( (n_1 - 1)s_1^2 = 12\times35^2=12\times1225 = 14700 \)

\( (n_2 - 1)s_2^2=13\times41^2 = 21853 \)

\( s_p=\sqrt{\frac{14700 + 21853}{25}}=\sqrt{1462.12}\approx38.238 \)

\( SE = 38.238\sqrt{\frac{1}{13}+\frac{1}{14}}\approx14.80 \)

\( t=\frac{160 - 181}{14.80}=\frac{-21}{14.80}\approx - 1.419 \)

But since the problem is incomplete, we need the actual values. However, the correct approach is:

1. For two - sample t - test with equal variances:

Test statistic formula:

\( t=\frac{(\bar{x}_1-\bar{x}_2)-d_0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \)

where \( d_0 = 0 \) (under null hypothesis of no difference), \( s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}} \)

2. Identify \( n_1 \), \( \bar{x}_1 \), \( s_1 \), \( n_2 \), \( \bar{x}_2 \), \( s_2 \) from the problem.

Assuming the missing values (from the original problem's likely context, let's say the sample from Location A: \( n_1 = 13 \), \( \bar{x}_1 = 150 \), \( s_1 = 35 \); sample from Location B: \( n_2 = 14 \), \( \bar{x}_2 = 181 \), \( s_2 = 41 \))

Step3: Recalculate with Correct Assumptions (if available)

Wait, the original problem's first part: "the stomach contents of a sample of 13 blue crabs from [Location A] of fish and a standard deviation of 35 milligrams. The stomach contents of [Location B] contain a mean of 181 milligrams of fish and a standard deviation of 41 milligrams." Let's assume \( n_1 = 13 \), \( \bar{x}_1 \) (let's say the mean for A is 160, maybe a typo in the original problem), \( n_2 \) (let's say 14, common sample size).

Then:

\( s_p=\sqrt{\frac{(13 - 1)\times35^2+(14 - 1)\times41^2}{13 + 14-2}}=\sqrt{\frac{12\times1225+13\times1681}{25}}=\sqrt{\frac{14700 + 21853}{25}}=\sqrt{\frac{36553}{25}}\approx\sqrt{1462.12}\approx38.238 \)

\( t=\frac{(160 - 181)-0}{38.238\sqrt{\frac{1}{13}+\frac{1}{14}}}=\frac{-21}{38.238\times\sqrt{\frac{27}{182}}}=\frac{-21}{38.238\times0.387}\approx\frac{-21}{14.80}\approx - 1.419 \)

But since the problem's text is cut, we need the exact values. However, if we assume that the mean for Location A is, for example, the value that was missing (maybe 150), then the calculation is as above.

But since the problem is incomplete, but to provide a solution, let's assume the correct values. Let's suppose the complete problem has:

Sample 1 (Location A): \( n_1 = 13 \), \( \bar{x}_1 = 150 \), \( s_1 = 35 \)

Sample 2 (Location B): \( n_2 = 14 \), \( \bar{x}_2 = 181 \), \( s_2 = 41 \)

Then the test statistic \( t=\frac{150 - 181}{\sqrt{\frac{(13 - 1)\times35^2+(14 - 1)\times41^2}{13 + 14-2}\times(\frac{1}{13}+\frac{1}{14})}} \)

First, calculate the pooled variance:

\( s_p^2=\frac{(13 - 1)\times35^2+(14 - 1)\times41^2}{13 + 14-2}=\frac{12\times1225 + 13\times1681}{25}=\frac{14700+21853}{25}=\frac{36553}{25}=1462.12 \)

Standard error \( SE=\sqrt{1462.12\times(\frac{1}{13}+\frac{1}{14})}=\sqrt{1462.12\times\frac{27}{182}}\approx\sqrt{1462.12\times0.14835}\approx\sqrt{216.9}\approx14.73 \)

\( t=\frac{150 - 181}{14.73}=\frac{-31}{14.73}\approx - 2.104 \)

But this is with assumed means. Since the problem's text is cut, we can't be sure, but the correct formula is used.

Answer:

(Assuming the correct values, the test statistic \( t\approx - 2.104 \) (rounded to three decimal places). But with the actual values from the complete problem, the calculation will be precise. If we assume the mean for Location A is 160, \( t\approx - 1.419 \), if mean is 150, \( t\approx - 2.104 \))

(Note: The problem's text is incomplete, so the above is a demonstration of the method. The actual answer depends on the missing mean value for Location A and the sample size for Location B.)