Answer to Question #234014 in Statistics and Probability for mira

Question

cyril randomly selected a sample of 100 children with albinism and found that only 70 of them are in special need schools. What is the upper confidence limit estimate of the proportion of children with albinism in special need schools? Consider 95% confidence level.

Answer 1

$$\hat{p}=\dfrac{X}{N}=\dfrac{70}{100}=0.7$$

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(0.7-1.96\sqrt{\dfrac{0.7(1-0.7)}{100}}, 0.7+1.96\sqrt{\dfrac{0.7(1-0.7)}{100}})$$

$$=(0.6102,0.7898)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.6102<p<0.7898,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.6102, 0.7898).$

The upper confidence limit estimate of the proportion of children with albinism in special need schools is 79.