Question

Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be profitable. Being very cautious, you set a goal of having 0.05 as the upper limit of a 90% confidence interval, when repeating the previous experiment. What should the maximum number of defective widgets be, out of 1024, for this goal to be reached.

Answer 1

Answer: 63 defective widgets

Step-by-step explanation:

Given that the proportion should not exceed 5%, that is:

p< or = 5%.

So we take p = 5% = 0.05

q = 1 - 0.05 = 0.095

Where q is the proportion of non-defective

We need to calculate the standard error (standard deviation)

S = √pq/n

Where n = 1024

S = √(0.05 × 0.095)/1024

S = 0.00681

Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.

UCL = p + Za/2(n-1) × S

Where a is alpha of confidence interval = 100 -90 = 10%

a/2 = 5% = 0.05

UCL = p + Z (0.05) × 0.00681

Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.

Z a/2 = 1.64

UCL = 0.05 + 1.64 × 0.00681

UCL = 0.0612

Since the UCL in this case is a measure of proportion of defective widgets

Maximum defective widgets = 0.0612 ×1024 = 63

Alternatively

UCL = p + 3√pq/n

= 0.05 + 3(0.00681)

= 0.05 + 0.02043 = 0.07043

UCL =0.07043

Max. Number of widgets = 0.07043 × 1024

= 72