Question

In: Statistics and Probability

Assume that the weights of trout in a river or normally distributed. You randomly catch and...

Assume that the weights of trout in a river or normally distributed. You randomly catch and weigh 40 such trout. The mean weight from the sample is 7.3 lb with a standard deviation of 0.9 lb. Test the claim that the mean weight of trout in this river is greater than 7lb at a significance level of 0.01 lb. Show your work.

Solutions

Expert Solution

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 7.0
Alternative hypothesis: u > 7.0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.1423
DF = n - 1

D.F = 39
t = (x - u) / SE

t = 2.08

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 2.08.

Thus the P-value in this analysis is 0.022.

Interpret results. Since the P-value (0.022) is greater than the significance level (0.01), we cannot reject the null hypothesis.

From the above test we do have sufficient evidence in the favor of the claim that that the mean weight of trout in this river is greater than 7lb at a significance level of 0.01.


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