Answer:
[tex]101-2.33\frac{11}{\sqrt{15}}=94.382[/tex]
[tex]101+2.33\frac{11}{\sqrt{15}}=107.618[/tex]
So on this case the 98% confidence interval would be given by (94.382;107.618)
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=101[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=11[/tex] represent the population standard deviation
n=15 represent the sample size
Solution to the problem
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)
The point estimate of the population mean is [tex]\hat \mu = \bar X =11[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that [tex]z_{\alpha/2}=2.33[/tex]
Now we have everything in order to replace into formula (1):
[tex]101-2.33\frac{11}{\sqrt{15}}=94.382[/tex]
[tex]101+2.33\frac{11}{\sqrt{15}}=107.618[/tex]
So on this case the 98% confidence interval would be given by (94.382;107.618)
