Question

For a normally distributed population, : with a standard deviation of 3.8

An article claims the mean life expectancy of wolves is greater than 7 years. David is a researcher and he decides to sample the population. He obtains the age of death of 30 wolves that have been studied and found the mean age of death to be 8.2.

Determine the test statistic that David should use to help make decisions about whether or not the mean lifespan of wolves is greater then 7 years, with 95% confidence.

Answer 1

Given:

Population mean (µ) = 7

Sample size (n) =30

Sample mean = 8.2

Population standard deviation (sigma)= 3.8

Confidence level = 95% = 0.95

The claim statement is, "the mean lifespan of wolves is greater than 7 years".

That is µ > 7.

Hence, the claim statement goes under the alternative hypothesis.

Hence, the Null hypothesis is, H0: µ <= 7

And the alternative hypothesis is, H1: µ > 7

Here, we have the population standard deviation.

So, we need to use the Z test for testing.

Now, let's find the test statistic.

That is, test statistic Z = 1.7297

Now, let's find the P-value.

For finding the P-value, we can use a technology like excel/Ti84 calculator or Z table.

The following excel command is used to find the P-value.

= 1 - NORMSDIST(Test statistic)

= 1 - NORMSDIST(1.7297)

You will get, P-value = 0.04185

We are given: Level of significance (alpha) = 0.05.

Following is the decision rule for making the decision.

If, P-value > α, then we fail to reject the null hypothesis.

If, P-value <= α, then we reject the null hypothesis.

Here, given α = 0.05

And P-value = 0.04185

That is P-value < α.

Hence, we reject the null hypothesis.

That is, there is sufficient evidence to support the claim that the mean lifespan of walves is greater than 7 years with 95% confidence.