Question

Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​margin of error​, and the confidence interval

Answer 1

Answer:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$

Step-by-step explanation:

Previous concepts

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".  

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".

$\bar X=10$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma$ represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

$X \sim N(\mu, \sigma)$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$  

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The lenght of the interval correspond to:

$8.4 = 2ME$

$ME= \frac{8.4}{2}= 4.2$

And since we know the margin of error we can find the limits for the confidence interval:

$Lower = 10 -4.2=5.8$

$Upper = 10 +4.2=14.2$