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posted by  jay-jay on 5/29/2008 6:38:22 PM  |  status: Closed  

Does my answer match your answer????????????????????

Course Textbook Chapter Problem
Statistics and Probability N/A N/A N/A
Question Details:
The owner of a marina would like to believe that more than 40% of the sailboat owners use their boats more than 6 times each summer. A random sample of 70 sailboat owners showed 42 used their boats more than 6 times each summer. Use the appropriate test to determine if there is reason for the marina owner to believe more than 40% of the sailboat owners use their boats more than 6 times each summer, at the 0.001 level. Also find the p- value.
      _
z = X -μ            60-40         = 20  =   2.64
      s/ √n       √(60) (40)       7.559
                          42
p= 1-.9959 (table value)= .004

There is no reason to believe that more than 40% of the sailboat owners use their boats more than 6 times each summer because the p value is less than the 0.01 confidence level.

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posted by Pilly on 5/29/2008 10:04:00 PM  |  status: Live
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jay-jay's comment:
"Thanks for your help!"
Response Details:

 
At the .01 level the critical value for the test statistic is -2.575
Performing a Z-test we get the value of the test statistic as
 
 for a p-value of .0003
 
Since p < .01 and  since Z > -2.575, we do not reject the null hypothesis and conclude that the percent of owners who use their boats more than 6 times each summer is greater than or equal to 40%
“There's no sense in being precise when you don't even know what you're talking about.”

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There is no philosophy which is not founded upon knowledge of the phenomena, but to get any profit from this knowledge it is absolutely necessary to be a mathematician.
Daniel Bernoulli 
                  
 
 
 
 
                         
 
 
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