Question

Good wood? A lab supply company sells pieces of Douglas fir 4 inches long and

1.5 inches square for force experiments in science classes. From experience, the strength of these pieces of wood follows a distribution with standard deviation 3000 pounds. You want to estimate the mean load needed to pull apart these pieces of wood to within 1000 pounds with 95% confidence. How large a sample is needed?

Short answer

The required sample is$35$

Step-by-step solution

Given information

Standard deviation$\left(\sigma \right)=3000$

Confidence level$=95\%$

Margin of error$\left(E\right)=1000$

The objective is to find out the sample size to have the margin error within 1000pounds at 95%the confidence level.

We know,

The formula to compute the sample size is:

$n={\left(\frac{z_{{\displaystyle \frac{\alpha }{2}}}\times \sigma }{E}\right)}^2$

According to the standard normal table, the value of $z_{\frac{a}{2}}$corresponding to the $95\%$confidence level is$1.96$

The sample size can be calculated as:

$$\begin{gathered} n={\left(\frac{z_{{\displaystyle \frac{\alpha }{2}}\times \sigma }}{E}\right)}^2 \\ ={\left(\frac{1.96\times 3000}{1000}\right)}^2 \\ =34.5744 \\ \approx 35 \end{gathered}$$

Therefore, the sample size is$35$