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joja [24]
3 years ago
15

A fast-food restaurant has an automated beverage machine that dispenses a set amount of liquid based on a size setting. Suppose

that on the medium setting, the standard deviation of the amounts dispensed is Ï=17 mL . The manager of the restaurant plans on taking a sample of n medium drinks to construct 99\% percent confidence interval for the mean amount of liquid dispensed. They want the margin of error to be no more 10 mL
Mathematics
1 answer:
SashulF [63]3 years ago
8 0

Answer:

n=(\frac{2.58(17)}{10})^2 =19.24 \approx 20

So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =17 represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got z_{\alpha/2}=2.58, replacing into formula (b) we got:

n=(\frac{2.58(17)}{10})^2 =19.24 \approx 20

So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence

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