In a sample of 77 southern cities, it was found that 49 had republican controlled city councils. Determine the UPPER bound of the 95% confidence interval for the proportion of southern cities with republican controlled city councils.
Solution
Given that sample size $n = 77$, observed $X = 49$.
Thus the sample proportion is
$\hat{p}=\frac{X}{n}=\frac{49}{77}=0.636$.
Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.
The estimate of the population proportion is $\hat{p} =\frac{X}{n} =\frac{49}{77}=0.636$.
$100(1-\alpha)$% 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
$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.636*(1-0.636)}{77}}\\ & =0.107. \end{aligned} $$
$95$% confidence interval estimate for population proportion is
$$ \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.636 - 0.107 & \leq p \leq 0.636 + 0.107\\ 0.5289 & \leq p \leq 0.7438. \end{aligned} $$
Thus, $95$% confidence interval estimate of the true proportion is $(0.5289,0.7438)$.
Thus, upper bound of the $95$% confidence interval estimate of the true proportion is $0.7438$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
