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

Answer:

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

Step-by-step 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:

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

$83.2-1.98\frac{6.4}{\sqrt{100}}=81.933$    

$83.2+1.98\frac{6.4}{\sqrt{100}}=84.467$

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