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 16 blue crabs from Location A contain a mean of 199 milligrams of fish and a standard deviation of 38 milligrams. The stomach contents of a sample of 7 blue crabs from Location B contain a mean of 188 milligrams of fish and a standard deviation of 41 milligrams. At =0.01, can you support the researcher's claim? Assume the population variances are equal and that both samples are from normal populations.

null and alt hypothesis?

t=? (round to three decimal places)

p-value=? (round to four decimal places)

state conclusion

Explanation / Answer

The statistical software output for this problem is:

Two sample T hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 = 2
HA : 1 > 2
(with pooled variances)

Hypothesis test results:

Hence,

Null and alternative hypotheses:

H0 : 1 = 2
HA : 1 > 2

t = 0.624

P - value = 0.2696

Conclusion:

P - value > 0.05 so we do not reject Ho. There is not sufficient evidence to conclude that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B.

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