Answer:
The 99% confidence interval is (3.0493, 3.4907).
We are 99% sure that the true mean of the students Perry score is in the above interval.
Step-by-step explanation:
Our sample size is 21.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
.
Then, we need to subtract one by the confidence level
and divide by 2. So:
![\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005](https://tex.z-dn.net/?f=%5Cfrac%7B1-0.99%7D%7B2%7D%20%3D%20%5Cfrac%7B0.01%7D%7B2%7D%20%3D%200.005)
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have ![T = 2.528](https://tex.z-dn.net/?f=T%20%3D%202.528)
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
![s = \frac{0.40}{\sqrt{21}} = 0.0873](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B0.40%7D%7B%5Csqrt%7B21%7D%7D%20%3D%200.0873)
Now, we multiply T and s
![M = 2.528*0.0873 = 0.2207](https://tex.z-dn.net/?f=M%20%3D%202.528%2A0.0873%20%3D%200.2207)
Then
The lower end of the interval is the mean subtracted by M. So:
![L = 3.27 - 0.2207 = 3.0493](https://tex.z-dn.net/?f=L%20%3D%203.27%20-%200.2207%20%3D%203.0493)
The upper end of the interval is the mean added to M. So:
![LCL = 3.27 + 0.2207 = 3.4907](https://tex.z-dn.net/?f=LCL%20%3D%203.27%20%2B%200.2207%20%3D%203.4907)
The 99% confidence interval is (3.0493, 3.4907).
Interpretation:
We are 99% sure that the true mean of the students Perry score is in the above interval.