Question

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)

Answer 1

`ME= 1.8808 * (80)/(√(400)) =7.5232`

Explanation:

Previous concepts

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".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma =80` 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:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

Since the Confidence is 0.94 or 94%, the value of
`\alpha=0.06` and
`\alpha/2 =0.03`, 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
`z_(\alpha/2)=1.8808`

The margin of error is given by:

`ME= 1.8808 * (80)/(√(400)) =7.5232`