An analyst from an energy research institute in California wishes to estimate the 90% confidence interval for the average price of unleaded gasoline in the state. In particular, she does not want the sample mean to deviate from the population mean by more than $0.05. What is the minimum number of gas stations that she should include in her sample if she uses the standard deviation estimate of $0.31, as reported in the popular press

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)

s represent the sample 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 =0.31 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.64(0.31)}{0.05})^2 =103.38 \approx 104[/tex]

So the answer for this case would be n=104 rounded up to the nearest integer