Test statistic and p-value

Question

In 2004, a random sample of 46 coyotes in a region of northern Minnesota showed an average age to be 2.05 years with a standard deviation of 0.82 years. However, it is thought that the overall population mean age of coyotes is 1.75 years. Test the hypothesis that coyotes in this region of northern Minnesota live longer than the average of 1. 75 years. What are the null and alternative hypotheses for this test?

Answer 1

Null Hypothesis, H0: µ (mean) age of coyotes ≤ 1.75 years

Alternative Hypothesis, Ha:µ (mean) age of coyotes > 1.75 years 

 

 

Since the standard deviation was computed from the sample of 46, we must use the t distribution where the appropriate degrees of freedom are n-1 = 45

 

The test statistic t =(xbar - µ)÷ (std÷√n) = (2.05 - 1.75)÷ (.82÷√46) =2.4813

 

Using a t table (actually an excel function) at 45 degrees of freedom yields a p-value of .0084

 

This means that we can reject the null at a significance level of .1, .05, .02, .01,