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 46 fish caught, the mean length was x = 18.5 inches, with estimated standard deviation s = 3.0 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.

Answer 1

Solution :

Given that,

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 19

Ha: 19

The test statistics,

t =( - )/ (s /n)

= ( 18.5 - 19 ) / ( 3 / 46 )

= -1.130

Critical value of  the significance level is α = 0.05, and the critical value for a left-tailed test is

= -1.679

Since it is observed that t = -1.130 > = -1.679, it is then concluded that the null hypothesis is fail to rejected.

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

P-value = 0.1322

The p-value is p = 0.1332 > 0.05, it is concluded that the null hypothesis is fail to rejected.

The conclusions obtained by using both methods are the same.