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At the Crown Interior Pizza Hat location in Faridabad, a report showed that 431 out of 778 orders were for delivery. What is the minimum sample size needed to have a 4 percentage point error for the proportion of orders that were carry-out?

*If you used a program on the graphing calculator or formula can you specify which ones please?

User Mathieug
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Answer:

Explanation:

To determine the minimum sample size needed to have a 4 percentage point error for the proportion of orders that were carry-out, we can use the following formula:

n = (Z^2 * p * q) / E^2

where:

n is the sample size we want to determine

Z is the z-score associated with the desired level of confidence (let's assume a 95% confidence level, so Z = 1.96)

p is the estimated proportion of orders that were carry-out (we don't know this yet, so we'll use 0.5 as a conservative estimate)

q is 1 - p (i.e., the estimated proportion of orders that were delivery)

E is the desired margin of error (in this case, 0.04)

Using the given information, we can estimate that the proportion of orders that were carry-out is:

p = 1 - 431/778 = 0.446

Then, plugging in the values into the formula, we get:

n = (1.96^2 * 0.446 * 0.554) / 0.04^2 ≈ 602.25

Rounding up to the nearest whole number, the minimum sample size needed is 603.

Note that I used the formula for sample size calculation based on a finite population, since the total number of orders is known (778). If we used the formula for an infinite population, the result would be very similar (around 604). I calculated this by hand, without using any calculator program, but you could use a calculator or a spreadsheet to perform the calculations more efficiently.

User Hytool
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