Question

The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today’s sample contains 14 defectives. Determine a 88% confidence interval for the proportion defective for the process today. Place your LOWER limit, rounded to 3 decimal places, in the first blank. For example, 0.123 would be a legitimate answer. Place your UPPER limit, rounded to 3 decimal places, in the second blank. For example, 0.345 would be a legitimate entry.

Answer 1

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 160, \pi = (14)/(160) = 0.088`

88% confidence level

So
`\alpha = 0.12`, z is the value of Z that has a pvalue of
`1 - (0.12)/(2) = 0.94`, so
`Z = 1.555`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.088 - 1.555\sqrt{(0.088*0.912)/(160)} = 0.053`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.088 + 1.555\sqrt{(0.088*0.912)/(160)} = 0.123`

The 88% confidence interval for the proportion of defectives today is (0.053, 0.123)