Respuesta :
Answer:
The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 +/- 0.010
= (0.090, 0.110)
Lower: 0.090
Upper: 0.110
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
Given that;
Proportion of disks which are not defective p' = 1328/1475 = 0.900
Proportion of disks which are defective p = 1 - p' = 1 - 0.900 = 0.100
p = 0.100
Number of samples n = 1475
Confidence interval = 80%
z value(at 80% confidence) = 1.28
Substituting the values we have;
0.100 +/- 1.28√(0.100(1-0.100)/1475)
0.100 +/- 1.28(0.007811334658)
0.100 +/- 0.009998508363
0.100 +/- 0.010
= (0.090, 0.110)
The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 +/- 0.010
= (0.090, 0.110)
Lower: 0.090
Upper: 0.110
