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The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate...

The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of α found in the one-tail area row. For a left-tailed test, the column header is the value of α found in the one-tail area row, but you must change the sign of the critical value t to −t. For a two-tailed test, the column header is the value of α from the two-tail area row. The critical values are the ±t values shown. Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is μ = 19 inches. However, a survey reported that of a random sample of 51 fish caught, the mean length was x = 18.5 inches, with estimated standard deviation s = 2.7 inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than μ = 19 inches? Use α = 0.05. Solve the problem using the critical region method of testing (i.e., traditional method). (Round the your answers to three decimal places.) test statistic = critical value = State your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the average fish length is less than 19 inches. Reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches. Fail to reject the null hypothesis, there is sufficient evidence that the average fish length is less than 19 inches. Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches. Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same? The conclusions obtained by using both methods are the same. We reject the null hypothesis using the traditional method, but fail to reject using the P-value method. We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.

Homework Answers

Answer #1

Sample size = n = 51

Sample mean = = 18.5

Standard deviation = s = 2.7

Claim: The average length of a trout caught in Pyramid Lake is less than μ = 19 inches.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Degrees of freedom = n - 1 = 51 - 1 = 50

Critical value = 1.684    ( Using t table)

Test statistic | t | <  critical vaue we fail to reject null hypothesis.

Conclusion:  Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.

---------------------------------------------------------------------------------------

By using the P-value method:

P-value = P(T < - 1.32) = 0.0960

P-value >0.05 we fail to reject the null hypothesis.

Conclusion:

Fail to reject the null hypothesis, there is insufficient evidence that the average fish length is less than 19 inches.

Yes, both conclusions are the same.

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