Question

Suppose the investigators had made a rough guess of 330 for the value of s before collecting data.

What sample size would be necessary to obtain an interval width of 50 ml for a confidence level of 95%?

Answer 1

`n=((1.960(330))/(25))^2 =669.36 \approx 670`

So the answer for this case would be n=670 rounded up to the nearest integer

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)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

`ME=z_(\alpha/2)(s)/(√(n))` (a)

And on this case sicne we want a confidence interval with a wisth of 50 ml then we have that ME =50/2=25 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)

We can use an estimator of the population deviation the sample deviation
`\hat \sigma = s = 330`

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(330))/(25))^2 =669.36 \approx 670`

So the answer for this case would be n=670 rounded up to the nearest integer