Answer :
Answer:
[tex]41.7-2.71\frac{3.86}{\sqrt{39}}=40.02[/tex]
[tex]41.7+2.71\frac{3.86}{\sqrt{39}}=43.38[/tex]
So on this case the 99% confidence interval would be given by (40.02;43.38)
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".
[tex]\bar X=41.7[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=3.86 represent the sample standard deviation
n=39 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=39-1=38[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,38)".And we see that [tex]t_{\alpha/2}=2.71[/tex]
Now we have everything in order to replace into formula (1):
[tex]41.7-2.71\frac{3.86}{\sqrt{39}}=40.02[/tex]
[tex]41.7+2.71\frac{3.86}{\sqrt{39}}=43.38[/tex]
So on this case the 99% confidence interval would be given by (40.02;43.38)