A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 98% confidence interval with an error of no more than 0.07. A consultant has informed them that a previous study found the mean to be 3.1 fast food meals per week and found the standard deviation to be 1.1. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

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=1.1[/tex] represent the population standard deviation

n represent the sample size

Solution to the problem

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 =0.07 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 formula to find it:"=-NORM.INV(0.01;0;1)", and we got [tex]z_{\alpha/2}=2.326[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337[/tex]

So the answer for this case would be n=1337 rounded up to the nearest integer