Transcript supplement6 - Ka

Should Antelope Coffee Inc. open a new
shop at Montana?*
Example7.4 of
Newbold and Carlson and Thorne, 6th edition
Ka-fu Wong & Nipun Sharma
University of Hong Kong
29 March 2007
*The ppt is a joint effort: Nipun Sharma discussed the Example with Dr. Ka-fu Wong on 28th
March 2007; Ka-fu explained the problem; Nipun drafted the ppt; Ka-fu revised it. Use it at
your own risks. Comments, if any, should be sent to [email protected].
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The problem at hand:
 Antelope Coffee Inc. is considering the possibility of
opening a coffee shop in Montana. Previous research
shows that a shop will be successful if the per capita
annual income > 60,000. The standard deviation is
known to be 5000.
 From a random sample of 36, the mean income was
62,300.
 Does this sample provide enough evidence to show that
the shop will be successful?
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Summarize the information and rewrite the question





Population Mean: 60,000
Standard Deviation: 5000
Sample Mean: 62,300
Sample size : 36
Standard deviation of the sample mean
= 5000/361/2 = 5000/6 = 833.33
 Sample mean of 62300 > Population mean of 60000
 Naturally we tend to conclude that the mean is > 60000,
but we know that there is a chance we will observe a
sample mean larger than or equal to 62300 even if the
true population mean is 60000 or lower.
 Is it rare to observe such sample mean when the true
population mean is 60000 or lower?
3
Is it rare to observe a sample mean that is larger
than or equal to 62300 when the true population
mean is 60000?
 Prob(m  62300 | m=60000)
=Prob((m-60000)/833.33  (62300-60000)/833.33)
=Prob(Z  2.76)
=0.00289
 Yes! It is rare to observe a sample mean that is larger
than or equal to 62300 when the true population mean is
60000.
 That is, it is unlikely that the population mean is 60000.
4
Is it rare to observe a sample mean that is larger
than or equal to 62300 when the true population
mean is 59999?
 Prob(m  62300 | m=59999)
=Prob((m-59999)/833.33  (62300-59999)/833.33)
=Prob(Z  2.7612)
=0.00288
 Yes! It is rare to observe a sample mean that is larger
than or equal to 62300 when the true population mean is
59999.
 That is, it is unlikely that the population mean is 59999.
 More unlikely than when the population mean is
60000.
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Is it rare to observe a sample mean that is larger
than or equal to 62300 when the true population
mean is 59998?
 Prob(m  62300 | m=59998)
=Prob((m-59998)/833.33  (62300-59998)/833.33)
=Prob(Z  2.7624)
=0.00287
 Yes! It is rare to observe a sample mean that is larger
than or equal to 62300 when the true population mean is
59998.
 That is, it is unlikely that the population mean is 59998.
 More unlikely than when the population mean is
59999.
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Concluding remarks
 That is, based on the sample information, it is very likely
that the population mean is larger than 60000.
 Opening a new coffee shop is very likely to be a success.
 What we really want to get is
 Prob(m < 60000 | m=62300) or
 Prob(m > 60000 | m=62300)
= 1- Prob(m < 60000 | m=62300)
 More generally, we are interested in
 Prob(a < m < b | m=62300)
 Materials in Chapter 8: Confidence Intervals.
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