Respuesta :
Answer:
a) [tex]27.3-1.714\frac{5.6}{\sqrt{24}}=25.34[/tex]
[tex]27.3+1.714\frac{5.6}{\sqrt{24}}=29.26[/tex]
So on this case the 90% confidence interval would be given by (25.34;29.26)
b) [tex]\frac{(23)(5.6)^2}{35.172} \leq \sigma^2 \leq \frac{(23)(5.6)^2}{13.091}[/tex]
[tex] 20.507 \leq \sigma^2 \leq 55.097[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 4.528 \leq \sigma \leq 7.423[/tex]
So the best option would be:
4.528<σ<7.423
c) For this case the 90% means that we have 90% of confidence that the true parameter [tex]\mu, \sigma[/tex] is between the limits calculated on the intervals
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=27.3[/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=24 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)
Part a
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=24-1=23[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a talel to find the critical value. The excel command would be: "=-T.INV(0.05,23)".And we see that [tex]t_{\alpha/2}=1.714[/tex]
Now we have everything in order to replace into formula (1):
[tex]27.3-1.714\frac{5.6}{\sqrt{24}}=25.34[/tex]
[tex]27.3+1.714\frac{5.6}{\sqrt{24}}=29.26[/tex]
So on this case the 90% confidence interval would be given by (25.34;29.26)
Part b
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,23)" "=CHISQ.INV(0.95,23)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=35.172[/tex]
[tex]\chi^2_{1- \alpha/2}=13.091[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(23)(5.6)^2}{35.172} \leq \sigma^2 \leq \frac{(23)(5.6)^2}{13.091}[/tex]
[tex] 20.507 \leq \sigma^2 \leq 55.097[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 4.528 \leq \sigma \leq 7.423[/tex]
So the best option would be:
4.528<σ<7.423
Part c
For this case the 90% means that we have 90% of confidence that the true parameter [tex]\mu, \sigma[/tex] is between the limits calculated on the intervals
