Does anyone know how to solve this statistics problem?

Toyota provides an option of a sunroof and side air bag package in its Corolla model; this package costs $1400 ($1159 invoice price). Assume that prior to offering this option package, Toyota wants to determine the percentage of Corolla buyers who would pay $1400 extra for the sunroof and side air bag. How many Corolla buyers must be surveyed if we want to be 95% confident that the sample percentage is within 4 percentage points of the true percentage for all Corolla buyers.





Can anyone help solve this problem?

Does anyone know how to solve this statistics problem?
The answer will vary, depending on the test that Toyota uses in the determination. However, for each of the tests, there is a table to determine population versus confidence level. My stat books are packed or I'd be able to give you a more definite answer.
Reply: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





in this question we are only concerned with the error term and the width of the interval w.





z * sqrt( (pHat * (1-pHat)) / n) = w





n = pHat * (1 - pHat) (w/z) ^ -2








We have no information on pHat so we will use the value of 1/2. this maximizes the value and if n is sufficient fo p = 0.5, n will be sufficient for all other values of p.





the z-score for a 95% confidence interval is 1.96








n = 0.5 * (1 - 0.5) * (0.04 / 1.96) ^ -2


n = 600.25





n must be integer valued. Always take the ceiling to make sure the size of the interval is correct.





n = 601.
Reply:We need to know the standard deviation of buyers who would pay $1,400 extra to answer this question.


The formula is:


n=[1.96*sd/error]^2


error=4


1.96 comes from the 95 % confidence level and normal distribution assumption. Plug the standard deviation into the equation to find the required sample size.