A car dealer wants to estimate the price its competitors charge for similar cars. The dealer wants to be 95% sure that the maximum margin of error is no larger than \$500. A pilot sample showed a standard deviation of \$3215. How large a sample does the dealer need to take?
Solution
The formula to estimate the sample size required to estimate the population mean is
$$ n =\bigg(\frac{z* \sigma}{E}\bigg)^2 $$
where $\sigma$ is the population stabdard deviation, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.
Given that the margin of error $E =500$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.
The population standard deviation is $\sigma = 3215$.
The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.
The minimum sample size required to estimate the mean is
$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{1.96*3215}{500}\bigg)^2\\ & =158.8306\\ &\approx 159. \end{aligned} $$
Thus, the sample of size $n=159$ will ensure that the $95$% confidence interval for the mean will have a margin of error $500$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
