An investigator wants to estimate caffeine consumption in high school students.
A)- How many students would be required to ensure that a 95% confidence interval estimate for the mean caffeine intake (measured in mg) is within 15 units of the true mean? Assume that the standard deviation in caffeine intake is 68mg

B)-How many students would be required to estimate the proportion of students who consume coffee? Suppose we want the estimate to be within 5% of the true proportion with 95% confidence. please show work.

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".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma[/tex] represent the population standard deviation

n represent the sample size

Part a

The margin of error is given by this formula:

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

And on this case we have that ME =15 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

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 [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.960(68)}{15})^2 =78.94 \approx 79[/tex]

So the answer for this case would be n=79 rounded up to the nearest integer

Part b

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

The margin of error for the proportion interval is given by this formula:

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a*)

And on this case we have that [tex]ME = 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a*) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b*)

Since we don't have any info provided we can assume [tex]\hat p =0.5[/tex]. And replacing into equation (b*) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16[/tex]

And rounded up we have that n=385