Question

Suppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for repair or replacement. The manufacturer wants to estimate the magnitude of the problem in order to design a quality control program. How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence

Answer 1

Answer:

80 telephones should be sampled

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

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

89% confidence level

So $\alpha = 0.11$, z is the value of Z that has a p-value of $1 - \frac{0.11}{2} = 0.945$, so $Z = 1.6$.

How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence?

n telephones should be sampled, an n is found when M = 0.09. We have no estimate for the proportion, thus we use $\pi = 0.5$

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

$0.09 = 1.6\sqrt{\frac{0.5*0.5}{n}}$

$0.09\sqrt{n} = 1.6*0.5$

$\sqrt{n} = \frac{1.6*0.5}{0.09}$

$(\sqrt{n})^2 = (\frac{1.6*0.5}{0.09})^2$

$n = 79.01$

Rounding up(as 79 gives a margin of error slightly above the desired value).

80 telephones should be sampled