Monday, May 11, 2009

Can you help with my statistics homework?

This is the problem from class.





The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza?





A) 0.115%26lt;p%26lt;0.260


B) 0.086%26lt;p%26lt;0.289


C) 0.102%26lt;p%26lt;0.273


D) 0.075%26lt;p%26lt;0.300





And this is what I am getting





N= 80, p hat = 15/80 = 0.1875, q hat = 1 – 0.1875 = 0.8125 and z sub α/2 = 2.58





0.1875 -2.58√(0.175*0.825/500) %26lt; p %26lt; 0.1875 -2.58√(0.175*0.825/500)





0.1875 – 0.045 %26lt; p%26lt; 0.1875 + 0.045





0.1425 %26lt; p %26lt; 0.2325





My answer dos not mach the options, what am I doing wrong?

Can you help with my statistics homework?
%26lt;%26lt;0.1875 -2.58√(0.175*0.825/500) %26lt; p %26lt; 0.1875 -2.58√(0.175*0.825/500)%26gt;%26gt;





Why are you dividing by 500 in the denominator? That should be the sample size, which is 80. I get a margin of error of .1126, which when added to phat gives us .3001. I'm going to go with (D) for the answer.


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