Question

When a truckload of apples arrives at a packing plant, a random sample of 150 apples are selected and examined for bruises and other defects. In reality, 12% of the apples on a particular truck are bruised or otherwise unsatisfactory. (a) How many standard errors away from 0.12 would you need to go to contain 89% of the sample proportions of bad apples you might expect to find? (3 decimal places) 1 (b) Suppose you were going to construct an 89% confidence interval from this population. What critical value should you use? (3 decimal places)

Answer 1

Solution :

Given that,

n = 150

Point estimate = sample proportion = = 0.12

1 - = 1 - 0.12 = 0.88

a) =  [p( 1 - p ) / n] = [(0.12 * 0.88) / 150 ] = 0.027

b) At 89% confidence level

= 1 - 89%

=1 - 0.89 =0.11

/2 = 0.055

Z/2 = Z0.055  = 1.598

Margin of error = E = Z / 2 *

= 1.598 * 0.027

= 0.043

A 89% confidence interval for population proportion p is ,

± E

= 0.12  ± 0.043

= ( 0.077, 0.163 )