Respuesta :
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
[tex]df = 21-1 = 20[/tex].
Then, we need to subtract one by the confidence level [tex]\alpha[/tex] and divide by 2. So:
[tex]\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005[/tex]
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 [tex]T = 2.528[/tex]
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
[tex]s = \frac{0.40}{\sqrt{21}} = 0.0873[/tex]
Now, we multiply T and s
[tex]M = 2.528*0.0873 = 0.2207[/tex]
Then
The lower end of the interval is the mean subtracted by M. So:
[tex]L = 3.27 - 0.2207 = 3.0493[/tex]
The upper end of the interval is the mean added to M. So:
[tex]LCL = 3.27 + 0.2207 = 3.4907[/tex]
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.
