Question

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.

Answer 1

a)
`n=((1.960(68))/(15))^2 =78.94 \approx 79`

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

b)
`n=(0.5(1-0.5))/(((0.05)/(1.96))^2)=384.16`

And rounded up we have that n=385

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` represent the population standard deviation

n represent the sample size

Part a

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (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:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (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
`z_(\alpha/2)=1.960`, replacing into formula (b) we got:

`n=((1.960(68))/(15))^2 =78.94 \approx 79`

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

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

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

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a*)

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

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b*)

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

`n=(0.5(1-0.5))/(((0.05)/(1.96))^2)=384.16`

And rounded up we have that n=385