A statistics practitioner took a random sample of 45 observations from a population whose standard deviation is 30 and computed the sample mean to be 106. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.
A. Estimate the population mean with 95% confidence.
B. Estimate the population mean with 95% confidence, changing the population standard deviation to 57.
C. Estimate the population mean with 95% confidence, changing the population standard deviation to 12.

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=106[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s=30 represent the sample standard deviation

n=45 represent the sample size

2) Part a

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=45-1=44[/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(0.025,44)".And we see that [tex]t_{\alpha/2}=2.015[/tex]

Now we have everything in order to replace into formula (1):

[tex]106-2.015\frac{30}{\sqrt{45}}=96.987[/tex]

[tex]106+2.015\frac{30}{\sqrt{45}}=115.011[/tex]

So on this case the 95% confidence interval would be given by (96.987;115.011)

3) Part b

If the popualtion standard deviation is known, then the confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

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]

Now we have everything in order to replace into formula (1):

[tex]106-1.96\frac{57}{\sqrt{45}}=89.346[/tex]

[tex]106+1.96\frac{57}{\sqrt{45}}=122.654[/tex]

So on this case the 95% confidence interval would be given by (89.346;122.654)

4) Part c

[tex]106-1.96\frac{12}{\sqrt{45}}=102.494[/tex]

[tex]106+1.96\frac{12}{\sqrt{45}}=109.506[/tex]

So on this case the 95% confidence interval would be given by (102.494;109.506)