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 = 18.5 inches, with estimated standard deviation s = 2.8 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	=

Answer 1

The provided sample mean is 18.5 and the sample standard deviation is s = 2.8, and the sample size is n = 51

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 19

Ha:: μ < 19

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is t_c = -1.676

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = -1.275 > t_c = -1.676, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1041, and since p = 0.1041 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is less than 19, at the 0.05 significance level.

test statistic	=	-1.275
critical value	=	-1.676