Question

The Park Manager at Fort Fisher State Park in North Carolina believes the typical park visitor spends at least 90 minutes in the park during summer months. A sample of 18 visitors during the summer months of 2011 revealed the mean time in the park was 96 minutes with a standard deviation of 12 minutes. At the 0.01 significance level, is it reasonable to conclude that the mean time in the park is greater than 90 minutes? What is alpha?

Answer 1

Given: Population mean (µ) = 90.

Sample size(n) = 18

Sample mean = xbar = 96.

Sample standard deviation = S = 12

The claim statement is, "the typical park visitor spends at least 90 minutes".

That is µ > 90.

Hence, the claim statement goes under the alternative hypothesis.

Hence, the Null hypothesis is, H0: µ <= 90

And the alternative hypothesis is, H1: µ > 90

Here, we have the sample standard deviation. So, we need to use the T-test for testing.

Now, let's find the test statistic.

That is, test statistic T = 2.121

Degrees of freedom = n-1 = 18 - 1 = 17 .

Now, let's find the P-value.

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

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

= 1 - T.DIST(Test statistic, Degrees of freedom, 1)

= 1 - T.DIST(2.121 , 17 , 1)

You will get, P-value = 0.0244

We are given: Level of significance = 0.01.

The level of significance is nothing but the α.

Hence α = 0.01.

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 = 0.01

And P-value = 0.0244

That is P-value > α.

Hence, we fail to the null hypothesis.

That is, there is not sufficient evidence to support the claim that the mean time in the spark is greater than 90 minutes.

That is, at the 0.01 significance level, it is not reasonable to conclude that the mean time in the park is greater than 90 minutes.

And alpha = 0.01.