Respuesta :
Answer:
Case [tex] s =7.2[/tex]
[tex]35.5-2.36\frac{7.2}{\sqrt{8}}=29.49[/tex]
[tex]35.5+2.36\frac{7.2}{\sqrt{8}}=41.51[/tex]
So on this case the 95% confidence interval would be given by (29.49;41.51)
Case [tex] \sigma =9.3[/tex]
[tex]35.5-1.96\frac{9.3}{\sqrt{8}}=29.06[/tex]
[tex]35.5+1.96\frac{9.3}{\sqrt{8}}=41.94[/tex]
So on this case the 95% confidence interval would be given by (29.06;41.94)
And we conclude that the intervals are very similar.
Step-by-step explanation:
If we assume that for this question we need to find a confidence interval for the population mean. We have the following procedure:
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= 35.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=7.2 represent the sample standard deviation
n=8 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=8-1=7[/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(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]
Now we have everything in order to replace into formula (1):
[tex]35.5-2.36\frac{7.2}{\sqrt{8}}=29.49[/tex]
[tex]35.5+2.36\frac{7.2}{\sqrt{8}}=41.51[/tex]
So on this case the 95% confidence interval would be given by (29.49;41.51)
If we assume that the real population standard deviation is [tex] \sigma =9.3[/tex] the confidence interval is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/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: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
[tex]35.5-1.96\frac{9.3}{\sqrt{8}}=29.06[/tex]
[tex]35.5+1.96\frac{9.3}{\sqrt{8}}=41.94[/tex]
So on this case the 95% confidence interval would be given by (29.06;41.94)
And we conclude that the intervals are very similar.