Question

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

Confidence intervals
ANSWERED
asked 2021-08-02
margin of error: $130,confidence level: \(\displaystyle{95}\%,\sigma=\${591}\) what is the sample size?

Answers (1)

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.
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