A mail-order company promises its customers that the products ordered will be ma

Question

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 38 of them were mailed within 72 hours of the placement of the orders. Construct a 98 % confidence interval for the percentage of all orders that are mailed within 72 hours of their placement.

Explanation / Answer

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=38
Sample Size(n)=50
Sample proportion = x/n =0.76
Confidence Interval = [ 0.76 ±Z a/2 ( Sqrt ( 0.76*0.24) /50)]
= [ 0.76 - 2.326* Sqrt(0.0036) , 0.76 + 2.33* Sqrt(0.0036) ]
= [ 0.6195,0.9005] ~ [0.62,0.9]
Interpretations:
1) We are 98% sure that the interval [0.6195 , 0.9005 ] contains the true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion  

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