Question

In: Statistics and Probability

A production line manager wants to determine how well the production line is running. He randomly...

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)

Solutions

Expert Solution

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)

orchestra answered 11 months ago
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