A researcher claims that the stomachs of blue crabs from Location A contain more

Question

A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 15 blue crabs from Location A contain a mean of 190 milligrams of fish and a standard deviation of 36 milligrams. The stomach contents of a sample of 8 blue crabs from Location B contain a mean of 182 milligrams of fish and a standard deviation of 44 milligrams. At alpha = 0.05, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below. (a) Identify the null and alternative hypotheses. Choose the correct answer below. A. H_0: mu_1 - mu_2 lessthanorequalto 0 H_a: mu_1 - mu_2 > 0 B. H_0: mu_1 - mu_2 greaterthanorequalto 0 H_a: mu_1 - mu_2

Explanation / Answer

(a) Null and ALternative Hypothesis

H0 : 1 - 2 <= 0

Ha :1 - 2 > 0

(b) Here we assume the population variance for both locations are same.

so Pooled standard deviation sp = sqrt [{(n1 -1) s12 + (n2 -1)s2 2 }/ (n1 + n2 -2)]

sp = sqrt [ (14 * 362 + 7 * 442 ) / 21] = 38.85

so Test statistic

t = (x1 - x2 ) / sp * sqrt [1/n1 + 1/n2 ] = (190 - 182)/ 38.85 * sqrt (1/15 + 1/8) = 8/ 17 = 0.470

P- value = 0.322

(D) DOn't reject the H0 . There is not a enough evidence to support researcher's claim.

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