There were  two of the best golfers to ever play the game. The
question was raised as to how these two golfers would have compared if
both were playing at the top of their game. The following sample data
show the results of 18-hole scores during a PGA tournament
competition. Wrigh's scores are from his 1960 season, while Good?s
scores are from his 1999 season (Golf Magazine,

February 2000). (Reference:  Anderson, Sweeney, Williams, 2002, p. 115)



Use the sample results to test the hypothesis of no difference between
the population mean 18-hole scores for the two golfers.  (The mean
value represents the numerical mean of the scores taken from the
rounds of golf played, represented by N)



                  Wright, 1960                          Goods, 1999

                  Mean = 69.95                          Mean = 69.56

                  N = 112                                   N = 84



a.       Assuming a population standard deviation of 2.5 for both
golfers, what is the value of the test statistic?

b.      What is the p-value?

c.       Using a = .01, what is your conclusion?



Write up your results, show the null and alternate hypotheses, set up
the problem and show the test statistic.

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Clarification of Question by wrench234-ga on 14 Mar 2005 19:16 PST

Reviewing need guidance to make sure on the right track need by 10PM EST 15 March

Dear Wrench,

Thank you for your question.  Several of the questions you have posted
deal with the area in statistics known as hypothesis testing.  Here is
a brief description of how to go about general hypothesis testing:

Eric W. Weisstein. "Hypothesis Testing." From MathWorld--A Wolfram Web Resource
http://mathworld.wolfram.com/HypothesisTesting.html

The hypothesis of ?no difference between? the two golfers is the null
hypothesis ? the result if the two golfers scores were entirely based
on chance.  Another way to state this is that statistically there is
no difference in the means for the two players.

To choose which test statistic to use, one must consider what we know
about the distributions.  Here?s another site that goes through some
examples of how to go through the process:

http://www.fordham.edu/economics/vinod/Hawch11.doc

Here?s an excerpt:  

?? (select statistic) The choice depends on answer to 3 questions. 
(i) Is [sigma] known? (ii) Is the distribution Normal?  (iii) is n
large??


In this problem, we?re given the standard deviation (2.5 for each), we
can assume that the distributions for each are normal (i.e. Gaussian
or bell-shaped), and that N is large.  N is ?large? generally if N >
30.  For more information on determining how large is large for N, you
may want to read about the Central Limit Theorem:
http://mathworld.wolfram.com/CentralLimitTheorem.html

Here?s some details on normal distributions:
http://mathworld.wolfram.com/NormalDistribution.html

For this combination, we should choose the two sample t-test for
independent measures.  Here?s more info on it:
http://leeds-faculty.colorado.edu/luftig/Past_Course_Websites/Research_Methods/LectureSlides-Text/Chapter9/CHAP_9.PPT

To calculate it, use

   t = (x1 ? x2) / sqrt(sp^2/n1 + sp^2/n2)

     Where sp^2 = ((n1-1)s1^2 + (n2-1)s2^2) / (n1 + n2 ?2)

Calculate sp^2 first:

Sp^2 = ((112-1)*2.5^2 + (84-1)*2.5^2) / (112+84-2)
        = (693.75 + 518.75) / 194
Sp^2 = 6.25

Now solve the first equation for the test statistic t:

t = (69.95 ? 69.56) / sqrt(6.25^2/112 + 6.25^2/84)
   = 0.39 / sqrt(0.34877 + 0.46503)
   = 0.39 / 0.9021
t  = 0.4323


The question asks for the p-value.  It also gives us an a-value
(usually called alpha), for the acceptance level ? the chance we?re
willing to take that we will reject the null hypothesis incorrectly. 
Here?s a page that puts these terms together:
http://mathworld.wolfram.com/AlphaValue.html


Calculate the p-value:

This can be done manually by integrating the t-distribution, however,
one usually looks up the t-value in a table.  A shortcut is to use a
t-value calculator.  Here?s a good one based on Java that also gives a
graph of the t-distribution for multiple degrees of freedom:
http://www.stat.sc.edu/~west/applets/tdemo.html

Using this tool, the p-value for t=0.4323 is 0.3701. 


Finally, for an alpha value of 0.01, p>alpha, so we must accept the
null hypothesis, i.e., we must conclude that there is no difference
between the two data sets and that there is no statistical difference
between the two golfers? records.


I hope this answer was helpful.  Feel free to ask for clarification.

         -welte-ga

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Request for Answer Clarification by wrench234-ga on 14 Mar 2005 20:54 PST

Thank you. I saw what I was doing thank you gain. I have one more if
you would like at it my calculations is off getting 6.45 it is under
PROBABILITY

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Clarification of Answer by welte-ga on 27 Mar 2005 11:51 PST

Glad to be of help - I'll take a look at the other statistics
questions if I have time.

Best,

         -welte-ga

Thank you I didnt raised the 2.5

Aha ... but the assumptions do not take account of the improvements in
club and ball technology between 1960 and 1999 ...

So, Mr Wright would demolish Mr Good, regardless of their p-values.