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

Answer:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$