Question

Explanation A random sample of 100 observations from a normally distributed population possesses a mean equal to 83.2 and a standard deviation equal to 6.4. Find a 95% confidence interval for μ.

Answer 1

The 95% confidence interval would be given by (81.933;84.467)

Explanation:

1) 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=83.2` represent the sample mean

`\mu` population mean (variable of interest)

s=6.4 represent the sample standard deviation

n=100 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

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

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=100-1=99`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that
`t_(\alpha/2)=1.98`

Now we have everything in order to replace into formula (1):

`83.2-1.98(6.4)/(√(100))=81.933`

`83.2+1.98(6.4)/(√(100))=84.467`

So on this case the 95% confidence interval would be given by (81.933;84.467)