Question

According to an independent research, a point estimate of the proportion of U.S. consumers of black tea is p = 0.76. Calculate the sample size needed to be 95% confident that the error in estimating the true value of p is less than 0.015? Use the z-value rounded to two decimal places to obtain the answer. 4072.69

Answer 1

Answer:

The sample size needed is 3115.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is:

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

Point estimate:

$\pi = 0.76$

95% confidence level

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

Calculate the sample size needed to be 95% confident that the error in estimating the true value of p is less than 0.015?

This is n for which M = 0.015. So

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

$0.015 = 1.96\sqrt{\frac{0.76*0.24}{n}}$

$0.015\sqrt{n} = 1.96\sqrt{0.76*0.24}$

$\sqrt{n} = \frac{1.96\sqrt{0.76*0.24}}{0.015}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.76*0.24}}{0.015})^2$

$n = 3114.26$

Rounding up:

The sample size needed is 3115.