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 +1.65)
12(1-12)
100
O 12 +1.964
12(1-12)
100
0.0.12 +1.65
0.12(1-0.12)
100
0.12(1 -0.12)
O 0.12 +1.967
100

0.12 ± 1.96 * √(0.12(0.88) / 100)

Step-by-step explanation:

Confidence interval :

Phat ± Zcritical * √(phat(1 -phat) / n)

Phat = 12/100 = 0.12

1 - phat = 0.88

Zcritical at 95% = 1.96

Hence, we have :

0.12 ± 1.96 * √(0.12(0.88) / 100)

0.12 ± 1.96 * 0.0324961

0.12 ± 0.0636924

Lower boundary = (0.12 - 0.0636924) = 0.0563

Upper boundary = 0.12 + 0.0636924 = 0.1837

there is no question content here, but you for the free points! : d

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.