Question

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.)

Answer 1

Answer:

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

In this problem, we have that:

$p = 0.55$

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 $M = 0.023$. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.023 = 1.96\sqrt{\frac{0.55*0.45}{n}}$

$0.023\sqrt{n} = 0.9751$

$\sqrt{n} = \frac{0.9751}{0.0023}$

$\sqrt{n} = 423.95$

$\sqrt{n}^{2} = (42.395)^{2}$

$n = 1797$

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.