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 = 2.15 years,
with sample standard deviation s = 0.79 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.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.
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.
Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 1.75
Alternative Hypothesis: μ > 1.75
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (2.15 - 1.75)/(0.79/sqrt(41))
t = 3.242
Rejection Region
This is right tailed test, for α = 0.01 and df = 40
Critical value of t is 2.423.
Hence reject H0 if t > 2.423
Reject the null hypothesis, there is sufficient evidence that the
average age of Minnesota coyotes is higher than 1.75 years.
P-value Approach
P-value = 0.0012
As P-value < 0.01, reject the null hypothesis.
The conclusions obtained by using both methods are the same.
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