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.

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:

[tex]\bar X=780[/tex] represent the sample mean

[tex]\sigma=40[/tex] represent the population standard deviation

n represent the sample selected (variable of interest)

[tex]\alpha=0.02[/tex] significance level

Confidence =0.98 or 98%

Me= 12 represent the margin of error required

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

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=2.33[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.33(40)}{12})^2 =60.32 \approx 61[/tex]

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