Answer:
The 99% confidence interval for this sample mean is between 2.895 ppb and 2.903 ppb.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 7 - 1 = 6
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995[/tex]. So we have T = 3.7074
The margin of error is:
M = T*s = 3.7074*0.001 = 0.004
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.899 - 0.004 = 2.895 ppb.
The upper end of the interval is the sample mean added to M. So it is 2.899 + 0.004 = 2.903 ppb.
The 99% confidence interval for this sample mean is between 2.895 ppb and 2.903 ppb.
