Question

Play Golf America program. The Professional Golf Association (PGA) and Golf Digest have developed the Play Golf America program, in which teaching professionals at participating golf clubs provide a free 10-minute lesson to new customers. According to Golf Digest, golf facilities that participate in the program gain, on average, \(2,400 in greens fees, lessons, or equipment expenditures. A teaching professional at a golf club believes that the average gain in greens fees, lessons, or equipment expenditures for participating golf facilities exceeds \)2,400.

a. In order to support the claim made by the teaching professional, what null and alternative hypotheses should you test?

b. Suppose you selectα = 0.05. Interpret this value in the words of the problem.

c. For α = 0.05, specify the rejection region of a large sample test.

Short answer

a. The null and the alternative hypotheses are H₀ : μ₀ = $2400 against H_a : μ > $2400

b. Testing α at 0.05 means there is a 5% risk that the researcher will decide that the mean gains are greater than $2400 when they are not.

c. {x: Z≥1.645}

Step-by-step solution

Given information

Participating golf facilities receive an average of $2,400. According to a teaching professional at a golf club, the average increase in greens fees, lessons, or equipment purchases for participating in facilities exceeds $2,400.

Specifying the null and the alternative hypothesis

a.

The null hypothesis is the assumed-true hypothesis, while the alternative hypothesis is the hypothesis that must be demonstrated with the data.

It is assumed that the mean of the gain is $2400, making it the null hypothesis. The researcher seeks to prove that the mean of the gain exceeds $2400, making it the alternative hypothesis.

i.e.

H₀ :μ₀ = $2400 againstH_a :μ > $2400

Interpretation

b.

In the context of hypothesis testing, α denotes the risk that will reject the null hypothesis when it is, in fact, true. In the context of this problem, in testing 0.05, there is a 5% risk that the researcher will decide that the mean gains are more significant than $2400 when they are not.

Specifying the rejection region

c.

For large sample tests, Z distribution can be used where

Z ~ N(0,1)

Let x be the critical value

Then,
P (Z> x) = 0.05

From the standard normal table,

x = 1.645

So, the critical region is

[x,∞) = [1.645,∞ )

={x: Z ≥ 1.645}