A fellow-student calculates a 99% confidence interval for a mean and gets (0.020,0.045). She says that there is a 99% chance that the true mean is between 0.020 and 0.045. Is she right? Why or why not?

No. Once the confidence interval is calculated, the probability that it contains the true mean  is either 1 or 0. So then is not appropiate says that the confidence level is a chance. The best conclusion for this case would be:

We have 99% confidence that the true mean for the variable of interest on this case is between (0.02; 0.045)

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".  

represent the sample mean for the sample  

population mean (variable of interest)  

s represent the sample standard deviation  

n represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:  

(1)  

In order to calculate the mean and the sample deviation we can use the following formulas:  

(2)  

(3)  

After apply the procedure  we got that the interval is (0.02; 0.045) and we need to analyze if the following conclusion is right or no:

She says that there is a 99% chance that the true mean is between 0.020 and 0.045. Is she right? Why or why not?

The best answer for this case would be:

No. Once the confidence interval is calculated, the probability that it contains the true mean  is either 1 or 0. So then is not appropiate says that the confidence level is a chance. The best conclusion for this case would be:

We have 99% confidence that the true mean for the variable of interest on this case is between (0.02; 0.045)