Question

A large shipping company wants each order to arrive within the delivery window they give to the customer. They plan on taking a random sample of orders to construct a one-sample z interval to estimate what proportion of all orders arrive within their delivery window. They'll use a confidence level of 95%, percent, and they don't want the margin of error to exceed 2 percentage points. Previous data suggests that about 90%, percent of orders arrive within their delivery window. If we assume that p=0.90, what is the smallest sample size required to obtain the desired margin of error?

Answer 1

Answer:

The smallest sample size required to obtain the desired margin of error is 865.

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:

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

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

If we assume that p=0.90, what is the smallest sample size required to obtain the desired margin of error?

The margin is n when $p = 0.90, M = 0.02$. So

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

$0.02 = 1.96\sqrt{\frac{0.9*0.1}{n}}$

$0.02\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.02}$

$\sqrt{n} = 29.4$

$\sqrt{n}^{2} = (29.4)^{2}$

$n = 865$

The smallest sample size required to obtain the desired margin of error is 865.

Answer 2

The smallest sample size required to obtain the desired margin of error is 865.

The given parameters are:

$p =0.90$ --- the proportion of orders that arrive within delivery window

$CI =95\%$ --- the confidence level

$E =2\%$ --- the margin of error

The margin of error is calculated as:

$E = z\sqrt{\frac{p (1 - p)}{{n}}$

The z score at 95% confidence level is 1.96.

So, the formula becomes

$2\% = 1.96 * \sqrt{\frac{0.90 (1 - 0.90)}{{n}}$

$0.02 = 1.96 * \sqrt{\frac{0.90 * 0.10}{{n}}$

Divide both sides by 1.96

$0.0102 = \sqrt{\frac{0.90 * 0.10}{{n}}$

Square both sides

$0.00010404 = \frac{0.90 * 0.10}{n}$

$0.00010404 = \frac{0.09}{n}$

Make n the subject

$n = \frac{0.09}{0.00010404}$

$n = 865.051903114$

Approximate

$n = 865$

Hence, the smallest sample size required to obtain the desired margin of error is 865.

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