If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that the difference between the sample estimate and the actual market share for its new product was no more than 9%, what sample size would be needed?

We need a sample size of least 119

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

Sample size needed

At least n, in which n is found when

We don't know the proportion, so we use , which is when we would need the largest sample size.

Rounding up

We need a sample size of least 119

the answer is 5/3x+7/5y+7