Question

A computer company wants to determine the proportion of defective computer chips from a day’s production. A quality control specialist takes a random sample of 100 chips from the day’s production and determines that there are 12 defective chips. Assuming all conditions are met, he constructs a 95% confidence interval for the true proportion of defective chips from a day’s production. What are the calculations for this interval?

12 plus-or-minus 1.65 StartRoot StartFraction 12 (1 minus 12) Over 100 EndFraction EndRoot
12 plus-or-minus 1.96 StartRoot StartFraction 12 (1 minus 12) Over 100 EndFraction EndRoot
0.12 plus-or-minus 1.65 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction EndRoot
0.12 plus-or-minus 1.96 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction EndRoot


answer d

Answer 1

To calculate the 95% confidence interval for the proportion of defective computer chips, we can use the following formula:
CI = p ± z*√(p(1-p)/n)
where:
CI is the confidence interval
p is the proportion of defective computer chips in the sample
z* is the critical value of the standard normal distribution at the 95% confidence level (which is 1.96)
n is the sample size
First, let's calculate the point estimate for the proportion of defective computer chips:
Point estimate = number of defective chips / sample size = 12/100 = 0.12
Next, let's calculate the margin of error:
Margin of error = z*√(p(1-p)/n) = 1.96 * √((0.12)(1-0.12)/100) ≈ 0.0616
Finally, we can construct the confidence interval by adding and subtracting the margin of error from the point estimate:
CI = 0.12 ± 0.0616 ≈ (0.0584, 0.1816)
Therefore, we can be 95% confident that the true proportion of defective computer chips from a day's production is between 5.84% and 18.16%.