91影视

The confidence level is 99%.

The margin of error is 3 percentage points.

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

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

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

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

a.

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

$$\begin{gathered} n=\frac{{\left(z_{\frac{伪}{2}}\right)}^2脳0.25}{{\left(E\right)}^2} \\ =\frac{{\left(2.575\right)}^2脳0.25}{{\left(0.03\right)}^2} \\ =1841.84 \\ 鈮�1842. \end{gathered}$$

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,

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

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

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.