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

Explanation:

Formula to find the margin of error :
`E=z^*\sqrt{(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=(15)/(62)\approx0.242`

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

Then,

`E=(1.96)\sqrt{(0.242(1-0.242))/(62)}`

`E=(1.96)\sqrt{(0.242(0.758))/(62)}`

`E=(1.96)\sqrt{(0.183436)/(62)}`

`E=(1.96)√(0.00295864516129)`

`E=(1.96)(0.0543934293945)`

`E=0.106611121613\approx0.107`

Hence, the margin of error = E= 0.107