Aresearcher wishes to estimate the number of households with two cars. how large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 5%? a previous study indicates that the proportion of households with two cars is 24%

And rounded up we have that n=281

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

And replacing into equation (b) the values from part a we got:

And rounded up we have that n=281

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total