Question

# Margin of error: $130,confidence level: 95\%,\sigma =$591 what is the sample size?

Confidence intervals
margin of error: \$130,confidence level: $$\displaystyle{95}\%,\sigma=\{591}$$ what is the sample size?

2021-08-04
Step 1
From the provided information,
Margin of error $$\displaystyle{\left({E}\right)}=\{130}$$
Confidence level $$\displaystyle={95}\%$$
Population standard deviation $$\displaystyle{\left(\sigma\right)}=\{591}$$
Step 2
The z value at $$\displaystyle{95}\%$$ confidence level from the standard normal table is 1.96.
The required sample size can be obtained as:
$$\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\sigma}}{{{E}}}}\right)}^{{{2}}}$$
$$\displaystyle={\left({\frac{{{1.96}\times{591}}}{{{130}}}}\right)}^{{{2}}}$$
$$\displaystyle={79.3963}$$
$$\displaystyle\approx{80}$$
Thus, the required sample size is 80.