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.

A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.11 years, with sample standard deviation s = 0.81 years. However, it is thought that the overall population mean age of coyotes is μ = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use α = 0.01. Solve the problem using the critical region method of testing (i.e., traditional method). (Round 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 age of Minnesota coyotes is higher than 1.75 years.

Fail to reject the null hypothesis, there is sufficient evidence that the average age of Minnesota coyotes is higher than 1.75 years.    

Fail to reject the null hypothesis, there is insufficient evidence that the average age of Minnesota coyotes is higher than 1.75 years.

Reject the null hypothesis, there is insufficient evidence that the average age of Minnesota coyotes is higher than 1.75 years.

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 P-value method, but fail to reject using the traditional method.    

We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.

Answer 1

Solution: Given information

number of random sample=n=41, sample average=$\bar x$=2.11, sample standard deviation= s = 0.81,

population mean=$\mu$=1.75 , significance level=$\alpha$=0.01

To test the coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years.

Hypothesis is formulated as

$H_0 : \mu=1.75$

$H_a : \mu>1.75$

Test statistic

$t=\frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}$

$t=\frac{2.11-1.75}{0.81/\sqrt{41}}$

t=2.846

Critical value:

$t_{n-1, \alpha}$=$t_{41-1, 0.01}$ = 2.423

Critical value= $t_c$ =2.423 -----------( from t table)

Decision:

Since it is observed that t=2.846> $t_c$=2.423 , it is then concluded that the null hypothesis is rejected.

Reject the null hypothesis, there is sufficient evidence that the average age of Minnesota coyotes is higher than 1.75 years.

P-value

To use the t table, you need to look at the row for degrees of freedom =df=41-1=40 and the values for which absolute value of the statistic = 2.846 lies in between 3.307 and 2.704.

Now look at the top of the table for " Area in one tail" and you will notice that these two corresponds 0.001 and 0.005.

Here we get a range for P-value. But our significance level $\alpha$=0.01 is out sight the range.

So we reject the null hypothesis( P-value<0.01).

The conclusions obtained by using both methods are the same.