A survey of 120 female freshman showed that 18 did not wish to work after marriage. Find the 95% confidence interval of the true proportion of females who do not wish to work after marriage.
Solution
Let $X$ denote the number of female freshman who did not wish to work after marriage.
Given that sample size $n = 120$, observed $X = 18$.
Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{18}{120}=0.15$.
Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.
The estimate of the population proportion is $\hat{p} =\frac{X}{n} =\frac{18}{120}=0.15$.
$100(1-\alpha)$% confidence interval for population proportion is
$$ \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned} $$
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.
The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.
Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.
The margin of error for proportions is
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.15*(1-0.15)}{120}}\\ & =0.064. \end{aligned} $$
$95$% confidence interval estimate for population proportion is
$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.15 - 0.064 & \leq p \leq 0.15 + 0.064\\ 0.0861 & \leq p \leq 0.2139. \end{aligned} $$
Thus, $95$% confidence interval estimate of the true proportion of females who do not wish to work after marriage is $(0.0861,0.2139)$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
