Answer :
Answer:
B. 3.414 +/-.00149
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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=0.001[/tex] represent the population standard deviation known
n=3 represent the sample size
Confidence interval
Since the population standard deviation is known, 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)
In order to calculate the mean we can use the following formula:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
The mean calculated for this case is [tex]\bar X=3.414[/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 tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]3.414-2.58\frac{0.001}{\sqrt{3}}=3.414 -0.00149=3.413[/tex]
[tex]3.414+2.58\frac{0.001}{\sqrt{3}}=3.414+0.00149=3.415[/tex]
So on this case the 99% confidence interval would be given by (3.413;3.415) The best option for this case is:
B. 3.414 +/-.00149