Question

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 = 18.4 inches, with estimated standard deviation s = 3.1 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? 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. The conclusions obtained by using both methods are the same.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 19
Alternative Hypothesis: μ < 19

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (18.4 - 19)/(3.1/sqrt(51))
t = -1.382

Rejection Region
This is left tailed test, for α = 0.05 and df = 50
Critical value of t is -1.676.
Hence reject H0 if t < -1.676

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

P-value Approach
P-value = 0.0866
As P-value >= 0.05, fail to reject null hypothesis.

yes, conclusions Are the same

The conclusions obtained by using both methods are the same.