An insurance company is interested in conducting a study to to estimate the population proportion of teenagers who obtain a driving permit within 1 year of their 16th birthday. a level of confidence of 99% will be used and an error of no more than .04 is desired. there is no knowledge as to what the population proportion will be. the size of sample should be at least

n=1041  or higher

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

2) 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 99% 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)  

Since we don't have a prior estimation for th proportion of interest, we can use this value as an estimation And replacing into equation (b) the values from part a we got:

And rounded up we have that n=1041  or higher.

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

step-by-step explanation:

i am not sure sorry : c