Respuesta :
Answer:
[tex]5.5-2.701\frac{1.1}{\sqrt{42}}=5.04[/tex]
[tex]5.5+2.701\frac{1.1}{\sqrt{42}}=5.96[/tex]
So on this case the 99% confidence interval would be given by (5.04;5.96)
And the best option would be:
a. 5.04 and 5.96
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=5.5[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=1.1 represent the sample standard deviation
n=42 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=42-1=41[/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(0.005,41)".And we see that [tex]t_{\alpha/2}=2.701[/tex]
Now we have everything in order to replace into formula (1):
[tex]5.5-2.701\frac{1.1}{\sqrt{42}}=5.04[/tex]
[tex]5.5+2.701\frac{1.1}{\sqrt{42}}=5.96[/tex]
So on this case the 99% confidence interval would be given by (5.04;5.96)
And the best option would be:
a. 5.04 and 5.96
