A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.09 years, with sample standard deviation s = 0.88 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.

(a) What is the level of significance?

Answer: The level of Significance is 0.01

Step-by-step explanation: As mentioned in the question that α = 0.01 which implies the significance level. This relates with 99% confidence level.

The significance level tells you the probability of rejecting a null hypothesis by the test when it is really true.

So, in this question the hypothesis would be:

H o : u ≤ 1.75

H α : u > 1.75

If we run the test we could conclude whether or not to reject the null hypothesis. Hence, we use the following formula (since population standard deviation is unknown):

t n - 1 = * - u / s / √n

t 41 - 1 = 2.09 - 1.75 / 0.88 / √41

t 40 = 2.473

the value of t statistics = 2.473

t40 > 2.473

Hence, we reject the null hypothesis.