Question

If the confidence level is 90%, find the Margin of sampling error. The population is normally distributed, the sample size is 15 and the sample mean is 75 and the std. dev is 5.

Answer 1

Answer:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$

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".

$\bar X= 75$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=5 represent the sample standard deviation

n=15 represent the sample size  

Solution to the problem

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,14)".And we see that $t_{\alpha/2}=1.761$

The margin of error is given by this formula:

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

And if we replace the values we got:

$ME = 1.761* \frac{5}{\sqrt{15}}=2.273$