Question

A engine part is built and shipped from a factory. The manager needs to estimate true proportion of parts (p) in the shipment that are mildly defective before the shipment goes out. He takes a sample of n=400n=400 parts and finds that 5% are mildly defective. He has 80% confidence that p lies in a confidence interval (a,b). What is the interval (a,b)?

Answer 1

Solution :

Given that,

n = 400

Point estimate = sample proportion = = 0.05

1 - = 1 - 0.05 = 0.95

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

Z/2 = Z0.10= 1.282

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.282 (((0.05 * 0.95) / 400)

= 0.014

A 80% confidence interval for population proportion p is ,

± E

= 0.05 ± 0.014

=( 0.036, 0.064 )

a = 0.036

b = 0.064