Historical data shows that 4% of the components produced at a certain manufacturing facility are defective. A particularly harsh labor dispute has recently been concluded, and management is curious about whether it will result in any change in this figure of 4%. A random sample of 500 items indicated 16 defectives (3.2%)
A. What is the Population variance
B. Construct a 95% Confidence Interval
C. Can we conclude that a change has occurred?
D. Assume that sample size increased to 2000/64 defects are found. Would you change your answer to C?
So, as this is a binomial issue, for
A, I understand that Var(Population) = (0.04)(0.96) = 0.0384
However, for question B, when I construct the confidence interval should it be between the two differences
Δ = 0.04-0.032 =0.008
Hence,
CI = 0.008 ± 1.96 * [(0.032/0.968)/500 +(0.04)(0.96)]
as a binomial confidence interval between two population proportions using independent samples?
or, should it rather be
CI = 0.008 ± 1.96 * [(0.032/0.968)/500]?
Much confused... any help will be greatly appreciated thankyou...
A. What is the Population variance
B. Construct a 95% Confidence Interval
C. Can we conclude that a change has occurred?
D. Assume that sample size increased to 2000/64 defects are found. Would you change your answer to C?
So, as this is a binomial issue, for
A, I understand that Var(Population) = (0.04)(0.96) = 0.0384
However, for question B, when I construct the confidence interval should it be between the two differences
Δ = 0.04-0.032 =0.008
Hence,
CI = 0.008 ± 1.96 * [(0.032/0.968)/500 +(0.04)(0.96)]
as a binomial confidence interval between two population proportions using independent samples?
or, should it rather be
CI = 0.008 ± 1.96 * [(0.032/0.968)/500]?
Much confused... any help will be greatly appreciated thankyou...