Question

Suppose a candy connoisseur opened a random bag of candy and found that 15 of the 62 candies were red. Using this information, determine the margin of error that would be used to construct a 95% confidence interval for the proportion of all candies that are red. Please give the answer in decimal form precise to three decimal places.

Answer 1

Answer: E= 0.107

Step-by-step explanation:

Formula to find the margin of error : $E=z^*\sqrt{\dfrac{p(1-p)}{n}}$

, where n= sample size .

p= sample proportion

z* = critical value.

Given : Sample size : n= 62

Number of red candies out of 62 =  15

Proportion of red candies = $p=\dfrac{15}{62}\approx0.242$

We know that , the critical value for 95% confidence = z*= 1.96  [Using z-table]

Then,

$E=(1.96)\sqrt{\dfrac{0.242(1-0.242)}{62}}$

$E=(1.96)\sqrt{\dfrac{0.242(0.758)}{62}}$

$E=(1.96)\sqrt{\dfrac{0.183436}{62}}$

$E=(1.96)\sqrt{0.00295864516129}$

$E=(1.96)(0.0543934293945)$

$E=0.106611121613\approx0.107$

Hence, the margin of error = E= 0.107