Question

How large a sample is needed in Question 1 if we wish to be 98% confident that our sample mean will be within 12 hours of the true mean? Use the standard deviation provided in Question 1.

Answer 1

n=61

Explanation:

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".

We assume the following info:

`\bar X=780` represent the sample mean

`\sigma=40` represent the population standard deviation

n represent the sample selected (variable of interest)

`\alpha=0.02` significance level

Confidence =0.98 or 98%

Me= 12 represent the margin of error required

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =12 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 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got
`z_(\alpha/2)=2.33`, replacing into formula (b) we got:

`n=((2.33(40))/(12))^2 =60.32 \approx 61`

So the answer for this case would be n=61 rounded up to the nearest integer