Question

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?

Answer 1

Answer:

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 $\pi$, 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 $Z = 1.96$.

Sample size needed

At least n, in which n is found when $M = 0.09$

We don't know the proportion, so we use $\pi = 0.5$, which is when we would need the largest sample size.

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

$0.09 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.09\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.09})^{2}$

$n = 118.6$

Rounding up

We need a sample size of least 119