Mathematics: 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,

Hmm, I think this can be solved by plugging in x=1 and x=3 into the function and finding the different between them...

A large shipping company wants each order to arrive

Ответ:

MoparorNocar061401

19.01.2022

Mathematics

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.

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