Question

Suppose a quality control engineer wishes to estimate the proportion of defective items in a production line. i. (4 points) To estimate the proportion within 0.02 and with 98% confidence, what is the minimum required sample size if the company currently has no idea about the

Answer 1

Answer:

The minimum sample size required is 3385.

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 $1 - \frac{\alpha}{2}$.

The margin of error is:

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

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

To estimate the proportion within 0.02 and with 98% confidence, what is the minimum required sample size if the company currently has no idea about the

The minimum sample size required is n.

We find n when $M = 0.02$

We don't know the true proportion, so we use $\pi = 0.5$, which is when the largest minimum sample size will be needed.

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

$0.02 = 2.327\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.02})^{2}$

$n = 3384.33$

Rounding up

The minimum sample size required is 3385.