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 14 blue crabs from Location A contain a mean of 199 milligrams of fish and a standard deviation of 36 milligrams. The stomach contents of a sample of 7 blue crabs from Location B contain a mean of 183 milligrams of fish and a standard deviation of 43 milligrams. At alphaequals0.05​, 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. Upper H 0​: mu 1minusmu 2greater than or equals0 Upper H Subscript a​: mu 1minusmu 2less than0 B. Upper H 0​: mu 1minusmu 2less than0 Upper H Subscript a​: mu 1minusmu 2equals0 C. Upper H 0​: mu 1minusmu 2less than or equals0 Upper H Subscript a​: mu 1minusmu 2greater than0 D. Upper H 0​: mu 1minusmu 2equals0 Upper H Subscript a​: mu 1minusmu 2not equals0 ​(b) Find the standardized test statistic for mu 1minusmu 2. tequals nothing ​(Round to three decimal places as​ needed.) ​(c) Calculate the​ P-value. Pequals nothing ​(Round to four decimal places as​ needed.) ​(d) State the conclusion. ▼ Fail to reject Reject Upper H 0. There ▼ is is not enough evidence at the 5​% level of significance to support the​ researcher's claim. Click to select your answer(s).

Answer 1

Given : For location A : , ,

For location B : , ,

The pooled estimate is ,

Let , and be the population mean for location A and location B respectively.

d.f.=degrees of freedom=14+7-2=19

(a) Hypothesis : Vs  

(b) The value of the test statistic is ,

(c) p-value= ; From excel "=TDIST(0.9,19,1)"

(d) Decision : Here , p-value=0.1897>

Therefore , fail to reject Ho.

Conclusion : Hence , there is not enough evidence at the 5% level of significance to support the researcher's claim.