Question

A microbrewery plans to conduct a study on beer consumption among men between the ages of 29 and 45. A pilot test indicated that men in this age category consume an average of 6 beers per week with a standard deviation of 2.3. Of the microbrewery seeks a 95% confidence level and a precision of 0.5 beers, what size sample should be used in the study

Answer 1

Answer:

$n=(\frac{1.960(2.3)}{0.5})^2 =81.28 \approx 82$

So the answer for this case would be n=82 rounded up to the nearest integer

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".

$\bar X=6$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=2.3 represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The margin of error is given by this formula:

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

And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$   (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (3) we got:

$n=(\frac{1.960(2.3)}{0.5})^2 =81.28 \approx 82$

So the answer for this case would be n=82 rounded up to the nearest integer