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?

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

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 . So

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

answer: no

step-by-step explanation:

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