91影视

The formula to find sample size is $\text{n}=\frac{1}{2}{\left(\frac{z_{\frac{伪}{2}}}{d}\right)}^2$, where d is the percentage error.

The sample size n can be determined by using the following formula,

$\mathit{n}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\left(\frac{z_{\frac{伪}{2}}}{d}\right)}^{\mathbf{2}}$

Where is critical value and d is the percentage difference.

For the given confidence level 98% corresponds to $伪=0.02鈥�鈥�鈥�鈥�\text{and}鈥�鈥�鈥�鈥�\frac{伪}{2}=0.01$.

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{伪}{2}}\right)=1-\frac{伪}{2} \\ =0.99 \end{gathered}$$

In the standard normal table for positive z score, find the value closest to 0.99, which is 0.9901, corresponding row value 2.3 and column values is 0.03; this corresponds to the z-score of 2.33, which is the critical value role="math" localid="1648122456490" ${\mathbf{\text{z}}}_{\mathbf{\text{0.01}}}$.

Thus, the critical value is $z_{0.01}=2.33$

For 15% percentage error, the sample size is calculated using the given formula,

$$\begin{gathered} \text{n}=\frac{1}{2}{\left(\frac{z_{\frac{伪}{2}}}{d}\right)}^2 \\ =\frac{1}{2}{\left(\frac{2.33}{0.15}\right)}^2 \\ =120.64 \\ =121 \end{gathered}$$

Therefore, with 121 sample values we are 98% confident that s is within 15% of $蟽$ .