Statistics help Inference for Proportions?

An automobile manufacturer would like to know what proportion of its customers are dissatisfied with the service received from their local dealer. The customer relations department will survey a random sample of customers and compute a 95% confidence interval for the proportion that are dissatisfied. From past studies, they believe that this proportion will be about 0.23. Find the sample size needed if the margin of error of the confidence interval is to be about 0.01.


n =





After taking this survey with the sample size above, suppose 15% of the sample say that they are dissatisfied. What is the margin of error of the 95% confidence interval?


MOE =





If you had no prior information about possible values for p, how large a sample would be needed to be certain that the margin of error was at most 0.01 with 95% confidence?


n =

Statistics help Inference for Proportions?
large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval.





For large sample confidence intervals for the proportion in this situation you have:





pHat ± z * sqrt( (pHat * (1-pHat)) / n)





where pHat is the sample proportion


z is the zscore for having α% of the data in the tails, i.e., P( |Z| %26gt; z) = α


n is the sample size





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in the first question we have to find the value of n for which





z * sqrt( (pHat * (1-pHat)) / n) %26lt; 0.01





the z-score for the 95% CI is z = 1.96





1.96 * sqrt( 0.23 * ( 1 - 0.23 ) / n ) %26lt; 0.01


sqrt(0.1771 / n ) %26lt; 0.005102041


0.1771 / n %26lt; 2.603082e-05


n %26gt; 0.1771 / 2.603082e-05


n %26gt; 6803.474





you need an integer value for n so n %26gt; 6804 for the margin or error to be 0.01 or smaller.





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b)





1.96 * sqrt( 0.15 * (1 - 0.15) / 6804 ) = 0.008484554





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c)





worse case is to have a proportion of 0.50. this maximizes the square root function. This is easy to prove in just a couple simple calculus steps.





1.96 * sqrt( 0.5 * 0.5 / n ) %26lt; 0.01


n %26gt; (0.5 * 0.5) / ((0.01 / 1.96) ^ 2)


n %26gt; 9604