1. 2. A researcher claims that the stomachs of blue crabs from Location A contai
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Question
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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 15 blue crabs from Location A contain a mean of 190 milligrams of fish and a standard deviation of 39 milligrams. The stomach contents of a sample of 10 blue crabs from Location B contain a mean of 187 milligrams of fish and a standard deviation of 44 milligrams. At = 0.01, can you support the researchers 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. .xA. Ho' -P220 OD. (b) Find the standardized test statistic for 1-2 t-0.179 (Round to three decimal places as needed.) (c) Calculate the P-value P = 0.4297 (Round to four decimal places as needed.) (d) State the conclusion Fail to reject Ho. There is not enough evidence at the 1% level of significance to support the researchers claimExplanation / Answer
Solution:-
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State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: A< B
Alternative hypothesis: A > B
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 17.18
DF = 23
t = [ (x1 - x2) - d ] / SE
t = 0.175
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 0.175. We use the t Distribution Calculator to find P(t > 0.175) = 0.4313
Therefore, the P-value in this analysis is 0.4313
Interpret results. Since the P-value (0.4313) is greater than the significance level (0.01), we cannot reject the null hypothesis.
Fail to reject the H0, There is not sufficient evidence to support the researcher's claim.
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