Question

The manager of a large supermarket took a random sample of

1400 egg cartons and found that 112 cartons had at least one

broken egg.

a. Find a point estimate for the proportion of all egg cartons

that have at least one broken egg.

Ans 10a______________

b. Estimate this proportion using 92% C.I.

Answer 1

Solution:

Given,

n = 1400 ....... Sample size

x = 112 .......no. of successes in the sample

a)

Let denotes the sample proportion.

     = x/n   = 112/1400 = 0.08

Point estimate is 0.08

b)

Our aim is to construct 92% confidence interval.

c = 0.92

= 1- c = 1- 0.92 = 0.08

  /2 = 0.04

= 1.751 ..(use z table)

Now , the margin of error is given by

E = /2 *  

= 1.751 * [0.08 *(1 - 0.08)/1400]

= 0.013

Now the confidence interval is given by

( - E)   ( + E)

(0.08 - 0.013)   (0.08 + 0.013)

0.067 0.093

Required 92% Confidence Interval is (0.067 , 0.093 )