Question

Finding Sample Size Instead of using Table 7-2 for determining the sample size required to estimate a population standard deviation s, the following formula can be used

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

where corresponds to the confidence level and d is the decimal form of the percentage error. For example, to be 95% confident that s is within 15% of the value of s, use $z_{\frac{\alpha }{2}}=1.96$ and d = 0.15 to get a sample size of n = 86. Find the sample size required to estimate s, assuming that we want 98% confidence that s is within 15% of $\sigma$.

Short answer

For 98% confidence level, the required sample size is 121.

Step-by-step solution

Given information

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

Describe the formula to determine the sample size

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

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

Where is critical value and d is the percentage difference.

Find the critical value  zα2

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

Mathematically,

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

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 $z_{0.01}=2.33$

Find the required sample size

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

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

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