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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 13 blue crabs from Location A contain a mean of 199 milligrams of fish and a standard deviation of 39 milligrams. The stomach contents of a sample of 9 blue crabs from Location B contain a mean of 189 milligrams of fish and a standard deviation of 44 milligrams. At alpha equals 0.01, can you support the​ researcher's claim? Assume the population variances are equal.

Required:
a. Identify the null and alternative hypotheses. Choose the correct answer below.
b. Find the standardized test statistic for ​(Round to three decimal places as​ needed.)
t=
c. Calculate the​ P-value.

Sagot :

Well I assume you have a TI-84 or something

a)  null is  μa = μb

alternate is  μa > μb

where μa = the true mean number of fish contrained in the stomach of blue crabs in location A

μb = the true mean number of fish contrained in the stomach of blue crabs in location B×

B) this is a 2 sample t-test

t= (xa-xb)/ [tex]\sqrt{s1^{2} /n1+ s2^{2} /n2}[/tex] so you find the t value and look up on the T- table

or you use your calculator

t= 0.549 ⇒ p= 0.295

since out p value of 0.295 is greater than our alpha level of 0.01, we have no evidence to conclude that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B.

you also might want to check the assumptions.

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