Respuesta :
Answer:
[tex]7.72-2.433\frac{2.5}{\sqrt{20}}=6.360[/tex]
[tex]7.72+2.433\frac{2.5}{\sqrt{20}}=9.080[/tex]
So on this case the 97.5% confidence interval would be given by (6.360;9.080)
Step-by-step explanation:
For this case we assume the deviation given as s =2.5 and the mean [tex]\bar = 7.72[/tex]
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=7.72[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=2.5 represent the sample standard deviation
n=20 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=20-1=19[/tex]
Since the Confidence is 0.975 or 97.5%, the value of [tex]\alpha=0.025[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.0125,19)".And we see that [tex]t_{\alpha/2}=2.433[/tex]
Now we have everything in order to replace into formula (1):
[tex]7.72-2.433\frac{2.5}{\sqrt{20}}=6.360[/tex]
[tex]7.72+2.433\frac{2.5}{\sqrt{20}}=9.080[/tex]
So on this case the 97.5% confidence interval would be given by (6.360;9.080)
