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Q10E

Expert-verifiedFound in: Page 398

Book edition
13th

Author(s)
James T. McClave, P. George Benson, Terry Sincich

Pages
888 pages

ISBN
9780134506593

**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 select α = 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.**

a. The null and the alternative hypotheses are *H*_{0} : *μ*_{0} = $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}

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.

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.

*H*_{0} : *μ*_{0} = $2400 against *H*_{a} : *μ* > $2400

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.

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}

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