How many integrated circuits must be randomly selected and tested for time to failure in order to estimate the mean time to failure? We want 95% confidence that the sample mean is within 2 hr of the population mean, and the population standard deviation is known to be 18.6 hours.
Solution
Given that the margin of error $E =2$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.
The population standard deviation is $\sigma = 18.6$.
The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.
The minimum sample size required to estimate the mean is
$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{1.96*18.6}{2}\bigg)^2\\ & =332.26\\ &\approx 333. \end{aligned} $$
Thus, the sample of size $n=333$ will ensure that the $95$% confidence interval for the mean will have a margin of error $2$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
