Respuesta :
Answer:
The 95% confidence interval would be given by (18.906;25.094)
Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.
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".
[tex]\bar X =22[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=5.6 represent the sample standard deviation
n=15 represent the sample size
2) 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=15-1=14[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/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(0.025,14)".And we see that [tex]t_{\alpha/2}=2.14[/tex]
Now we have everything in order to replace into formula (1):
[tex]22-2.14\frac{5.6}{\sqrt{15}}=18.906[/tex]
[tex]22+2.14\frac{5.6}{\sqrt{15}}=25.094[/tex]
So on this case the 95% confidence interval would be given by (18.906;25.094)
Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.
