Question

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Lefties: Find the sample size needed to estimate the percentage of California residents who are left-handed. Use a margin of error of three percentage points, and use a confidence level of 99%.

a. Assume that $\hat{p}and\hat{q}$are unknown.

b. Assume that based on prior studies, about 10% of Californians are left-handed.

c. How do the results from parts (a) and (b) change if the entire United States is used instead of California?

Short answer

a. The sample size when both $\hat{p}and\hat{q}$ are unknown is 1842.

b. The sample size when 10% of Californians are left-handed is 664.

c. The results from part a and part b will not change.

Step-by-step solution

Given information

The confidence level is 99%.

The margin of error is 3 percentage points.

Requirements for determining sample size

The basic requirement is that the sample should be independent and randomly selected. In this case, this requirement has been satisfying.

Formulae for determining sample sizes

The sample size can be determined with 2 different conditions. The formulas and the conditions are given below.

a. When $\hat{\mathbf{p}}$is unknown, $\mathbf{n}\mathbf{=}\frac{{\left({\mathbf{z}}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}\right)}^{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{25}}{{\mathbf{E}}^{\mathbf{2}}}$

b. When $\hat{p}$is known $n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^2\times \hat{p}\hat{q}}{E^2}$

Find Critical value

The critical value $z_{\frac{\alpha }{2}}$is obtained from the standard normal table at 99% level of confidence. Now for 99% confidence level, the z-score will be

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{\frac{0.01}{2}} \\ =z_{0.005} \\ =2.575 \end{gathered}$$

Find sample size 

a.

As the value of sample proportion is unknown, the sample size is determined as follows,

$$\begin{gathered} n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^2\times 0.25}{{\left(E\right)}^2} \\ =\frac{{\left(2.575\right)}^2\times 0.25}{{\left(0.03\right)}^2} \\ =1841.84 \\ \approx 1842. \end{gathered}$$

Therefore when there is no other prior information, the sample size is 1842.

b.

As the value of sample proportion is known to be 0.10$\left(\hat{p}\right)$ , the sample size is determined as follows,

$$\begin{gathered} n=\frac{{\left(z_{\frac{\alpha }{2}}\right)}^2\times \hat{p}\times \hat{q}}{{\left(E\right)}^2} \\ =\frac{{\left(2.575\right)}^2\times 0.10\times \left(1-0.10\right)}{{\left(0.03\right)}^2} \\ =\frac{{\left(2.575\right)}^2\times 0.10\times 0.90}{{\left(0.03\right)}^2} \\ \approx 664 \end{gathered}$$

Therefore, the sample size when 10% Californians are left handed is 664.

Analyze the result 

c.

The results in each of the two subparts would not change as the values used in the computation are unaffected by population size.

It only depends on the margin of error, sample proportion, and type of testing conducted.