A research group conducted an extensive survey of 2968 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1631 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage? (Hint: Use p ≈ 0.55 as a preliminary estimate. Round your answer up to the nearest whole number.)

We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.

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 .

In this problem, we have that:

How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage?

We have to find n for which . So

We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.

i wish i could you, i have no idea how to answer this

step-by-step explanation:

it is 32.5

i hope i could u.