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

`ME = 1.761* (5)/(√(15))=2.273`

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)(s)/(√(n))`

And if we replace the values we got:

`ME = 1.761* (5)/(√(15))=2.273`