Question

14. In a random sample of 45 Skittles candies, 12 are the color yellow. Use the Agresti-Coull approach to find a 95% confidence interval for the proportion of yellow Skittles in existence. Report the end points of the confidence interval to three decimal places. Then, interpret the confidence interval in context (beginning with the phrase, "We are 95% confident that ").

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 12 / 45 = 0.2667

1 - $\hat p$ = 0.7333

Z $\alpha$ /2 = 1.96

Margin of error = E = Z $\alpha$ / 2 * $\sqrt$ (( $\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 * $\sqrt$ [(0.2667 * 0.7333) / 45]

= 0.127

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.2667 - 0.127 < p < 0.2667 - 0.127

0.1397 < p < 0.3937

We are 95% confidence that interval for the proportion of yellow Skittles in existance is : (0.140, 0.394)