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a production line manager wants to determine how well the production line is running. he randomly selected 90 items from 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 to estimate the population proportion of items that are defective.



Answer :

contexto1024

The 99% confidence interval to estimate the population proportion of items that are defective is (0.0116, 0.1662).

Sample size , n = 90

defective items , x = 8

Sample proportion = x / n

= 8/90 = 0.0889

= 1- 0.0889 =0.9111

At 99% confidence level the z is ,

α = 1 - 0.99 = 0.01 , α/ 2 = 0.01 / 2

= 0.005

Zα/₂ = Z0.005 = 2.576 ( Using z table )

Margin of error Formula, E = Zα/2× (sqrt(((p-hat× (1 - p-hat)) /n)

= 2.575 ×sqrt((0.0889*0.9111) /90 )

E = 0.0773

A 99% confidence interval is

p-hat - E < p < p-hat + E

=> 0.0889 - 0.0773 < p < 0.0889 +0.0773

=> 0.0116 < p< 0.1662

Hence, confidence interval is (0.0116, 0.1662).

To learn more about Confidence interval, refer:

https://brainly.com/question/15712887

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