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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 160, \pi = \frac{14}{160} = 0.088$

88% confidence level

So $\alpha = 0.12$, z is the value of Z that has a pvalue of $1 - \frac{0.12}{2} = 0.94$, so $Z = 1.555$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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