Question

a researcher claims that the stomachs of blue crabs from location a contain more fish than the sto from location b. the stomach contents of a sample of 13 blue crabs from location a contain a me of fish and a standard deviation of 35 milligrams. the stomach contents of a sample of 7 blue crab contain a mean of 181 milligrams of fish and a standard deviation of 41 milligrams. at α = 0.10, can the researcher’s claim? assume the population variances are equal. complete parts (a) through (d (a) identify the null and alternative hypotheses. choose the correct answer below. a. $h_0: mu_1 - mu_2 geq 0$ $h_a: mu_1 - mu_2 < 0$ b. $h_0: mu_1 - mu_2 = 0$ $h_a: mu_1 - mu_2
eq 0$ c. $h_0: mu_1 - mu_2 < 0$ $h_a: mu_1 - mu_2 = 0$ d. $h_0: mu_1 - mu_2 leq 0$ $h_a: mu_1 - mu_2 > 0$ (b) find the standardized test statistic for $mu_1 - mu_2$. $t = square$ (round to three decimal places as needed.)

Explanation:

Step1: Identify given values

Let Location A be population 1, Location B be population 2.
$n_1 = 13$, $\bar{x}_1 = 210$ (assuming the mean for Location A was cut off, likely 210 as a common value here), $s_1 = 35$
$n_2 = 7$, $\bar{x}_2 = 181$, $s_2 = 41$

Step2: Calculate pooled variance

Pooled variance formula: $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2 - 2}$
$(n_1 - 1)s_1^2=(13 - 1)35^2 = 12\times1225 = 14700$
$(n_2 - 1)s_2^2=(7 - 1)41^2 = 6\times1681 = 10086$
$n_1 + n_2 - 2 = 13 + 7 - 2 = 18$
$s_p^2=\frac{14700 + 10086}{18}=\frac{24786}{18}=1377$

Step3: Calculate standard error

Standard error formula: $SE = s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
$s_p=\sqrt{1377}\approx37.108$
$\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{13}+\frac{1}{7}=\frac{7 + 13}{91}=\frac{20}{91}\approx0.2198$
$SE = 37.108\times\sqrt{0.2198}\approx37.108\times0.4688\approx17.40$

Step4: Calculate t - statistic

t - statistic formula: $t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1 - \mu_2)}{SE}$
Null hypothesis: $\mu_1 - \mu_2 = 0$, so $t=\frac{(210 - 181)-0}{17.40}=\frac{29}{17.40}\approx1.667$ (Wait, if $\bar{x}_1$ was actually 200, let's recalculate. Wait, maybe the original mean for A is 200? Wait, the text says "a mean of fish" (cut off), likely 200. Let's correct: if $\bar{x}_1 = 200$:
$t=\frac{(200 - 181)-0}{SE}$, $SE$ as above. Wait, no, let's check again. Wait, the user's image has a typo, but assuming $\bar{x}_1 = 210$ (common in such problems). Wait, maybe the correct mean for A is 210. Then:
$\bar{x}_1 - \bar{x}_2 = 210 - 181 = 29$
$t=\frac{29}{17.40}\approx1.667$? Wait, no, wait $SE$ calculation: $\sqrt{\frac{1}{13}+\frac{1}{7}}=\sqrt{\frac{20}{91}}\approx\sqrt{0.2198}\approx0.4688$, $s_p = \sqrt{1377}\approx37.108$, so $SE = 37.108\times0.4688\approx17.40$. Then $t=\frac{29}{17.40}\approx1.6667\approx1.667$. Wait, but maybe the mean for A is 200. Let's recalculate with $\bar{x}_1 = 200$:
$\bar{x}_1 - \bar{x}_2 = 200 - 181 = 19$
$t=\frac{19}{17.40}\approx1.092$. Wait, this is confusing. Wait, the original problem: "the stomach contents of a sample of 13 blue crabs from Location A contain a mea[cut off] of fish and a standard deviation of 35 milligrams". Likely the mean is 200 (common). Wait, maybe I made a mistake in pooled variance. Wait, let's recalculate pooled variance:
$(13 - 1)35^2 = 12*1225 = 14700$
$(7 - 1)41^2 = 6*1681 = 10086$
Sum: 14700 + 10086 = 24786
df: 13 + 7 - 2 = 18
$s_p^2 = 24786 / 18 = 1377$, correct. $s_p = \sqrt{1377} \approx 37.108$, correct.
Standard error: $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} = 37.108*\sqrt{\frac{1}{13}+\frac{1}{7}} = 37.108*\sqrt{\frac{20}{91}} \approx 37.108*0.4688 \approx 17.40$, correct.
Now, if the mean for A is 210 (as a common value when such problems have 210), then $\bar{x}_1 - \bar{x}_2 = 210 - 181 = 29$. Then $t = 29 / 17.40 \approx 1.667$. But maybe the mean is 200. Wait, perhaps the original problem had $\bar{x}_1 = 200$. Let's check with $\bar{x}_1 = 200$: $200 - 181 = 19$, $t = 19 / 17.40 \approx 1.092$. But this is ambiguous. Wait, maybe the user's problem has $\bar{x}_1 = 210$ (since the claim is more fish in A, so A's mean should be higher). So proceeding with $\bar{x}_1 = 210$:

Answer:

$t\approx1.667$ (If $\bar{x}_1 = 200$, $t\approx1.092$; but likely $\bar{x}_1 = 210$, so $t\approx1.667$)