Question

a researcher claims that the stomachs of blue crabs from location a contain more from location b. the stomach contents of a sample of 13 blue crabs from location of fish and a standard deviation of 35 milligrams. the stomach contents of a sample contain a mean of 181 milligrams of fish and a standard deviation of 41 milligrams. a the researcher’s claim? assume the population variances are equal. complete parts
(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 = 0.690 (round to three decimal places as needed.)
(c) calculate the p - value.
p = 0.2496 (round to four decimal places as needed.)
(d) state the conclusion.
$h_0$. there enough evidence at the 10% level of significance t

Explanation:

Step1: Analyze Hypotheses (Part a)

The researcher claims Location A crabs have more fish (so \( \mu_1 > \mu_2 \), where \( \mu_1 \) is Location A mean, \( \mu_2 \) Location B). Null hypothesis \( H_0 \) is the opposite or equality, alternative \( H_a \) is the claim. So \( H_0: \mu_1 - \mu_2 \leq 0 \), \( H_a: \mu_1 - \mu_2 > 0 \), which is option D.

Step2: Standardized Test Statistic (Part b)

For pooled t - test (equal variances), formula: \( t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1 - \mu_2)_0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \), where \( s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \). Assume \( n_1 = 13 \), \( \bar{x}_1 \) (missing but from context, let's say \( \bar{x}_1 \) and \( \bar{x}_2 = 181 \), \( s_1 = 35 \), \( s_2 = 41 \), \( n_2 \) (missing, but calculation gives \( t = 0.690 \)).

Step3: P - value (Part c)

For right - tailed test with \( df=n_1 + n_2 - 2 \), using t - distribution, P - value is \( P(T > t) \). With \( t = 0.690 \) and appropriate df, P - value is 0.2496.

Step4: Conclusion (Part d)

At 10% significance, \( \alpha = 0.1 \). P - value (0.2496) > \( \alpha \), so we fail to reject \( H_0 \). There is not enough evidence.

Answer:

(a) D. \( H_0: \mu_1 - \mu_2 \leq 0 \), \( H_a: \mu_1 - \mu_2 > 0 \)
(b) \( t = 0.690 \)
(c) \( P = 0.2496 \)
(d) Fail to reject \( H_0 \); There is not enough evidence.