Student Question

# Suppose that you want to test the claim that the mean wingspan of the Eurasian eagle-owl is 64 inches or more. You collect a sample of 54 Eurasian eagle-owls and find that the mean wingspan in the sample is 62 inches. The standard deviation is 8 inches. Using a level of significance α of 0.05, test the claim using the p-value.

We are asked to test the claim that the average wing span is greater than or equal to 64 inches. We have a sample of size 54 with a mean of 62 inches. The population standard deviation is 8 inches. We are asked to test at the 95% confidence level.

` H_0: mu=64 ` This is the null hypothesis and the claim.

`H_1: mu < 64 ` This is the alternative hypothesis.

Since the alternative hypothesis is one-tailed, we find the critical value z such that the area to the left is .05. From a standard normal table or technology we find the critical value to be -1.645 and the critical region to be z<-1.645. (Some texts will use 1.64 or 1.65 here.)

We compute the test value: `z= (62-64)/(8/sqrt(54))~~-1.837 `

Using a standard normal table or technology we get `p~~0.033 ` (My calculator gives `p~~0.0330962301 ` )

Since p=.033<.05 we reject the null hypothesis.

There is sufficient evidence to reject the claim that the average wing span is 64 or more inches.