Question

12)

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 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.05 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 insufficient 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 sufficient 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?

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

12.
Given that,
population mean(u)=1.75
sample mean, x =2.05
standard deviation, s =0.81
number (n)=51
null, Ho: μ=1.75
alternate, H1: μ>1.75
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.403
since our test is right-tailed
reject Ho, if to > 2.403
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.05-1.75/(0.81/sqrt(51))
to =2.645
| to | =2.645
critical value
the value of |t α| with n-1 = 50 d.f is 2.403
we got |to| =2.645 & | t α | =2.403
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 2.645 ) = 0.00544
hence value of p0.01 > 0.00544,here we reject Ho
ANSWERS
---------------
null, Ho: μ=1.75
alternate, H1: μ>1.75
test statistic: 2.645
critical value: 2.403
decision: reject Ho
p-value: 0.00544
we have enough evidence to support the claim that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years.

p value method
given data,
test statistic: 2.645,
sample size is 51
level of significance =0.01
Critical Value
The Value of t α at 0.01 LOS with n-1 = 50 is +2.4033          
P-Value : Right Tail - Ha :( P > 2.645) = 0.005444          
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 2.645 ) = 0.00544
hence value of p0.01 > 0.00544,here we reject Ho