In: Statistics and Probability
A production line manager wants to determine how well the production line is running. He randomly selected 90 items off of the assembly line and found that 8 were defective. (Assume all conditions have been satisfied for building a confidence interval). Find the 99% confidence interval.
| 
 (0.0234, 0.1099)  | 
|
| 
 (0.0116, 0.1662)  | 
|
| 
 (0.0301, 0.1477)  | 
|
| 
 (0.0396, 0.1382)  | 
Solution :
Given that,
n = 90
x = 8
Point estimate = sample proportion = 
 = x / n = 8/90=0.0889
1 - 
  = 1- 0.0889 =0.9111
At 99% confidence level the z is ,
  = 1 - 99% = 1 - 0.99 =
0.01
/ 2 = 0.01 / 2 = 0.005
Z
/2
= Z0.005 = 2.576 ( Using z table )
  Margin of error = E = Z
/2   *
((((
* (1 - 
)) / n)
= 2.576* (((0.0889*0.9111)
/90 )
E = 0.0773
A 99% confidence interval is ,
- E < p < 
 + E
0.0889 - 0.0773 < p < 0.0889 +0.0773
0.0116 , 0.1662
| 
 (0.0116, 0.1662)  |