A population has a standard deviation of 80. A random sample of 400 items from this population is selected. The sample mean is determined to be 200. Find the margin of error at 94% confidence level assuming that the population is large. (Round your solution to 4 decimal places)

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 =80[/tex] represent the population standard deviation

n=400 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

Since the Confidence is 0.94 or 94%, the value of [tex]\alpha=0.06[/tex] and [tex]\alpha/2 =0.03[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that [tex]z_{\alpha/2}=1.8808[/tex]

The margin of error is given by:

[tex] ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232[/tex]