Answer:
a) The 95% confidence interval to estimate μ is (7.435, 8.045) hours.
b) The 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam is (41.079%, 45.481%).
Step-by-step explanation:
Question a:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 1.96\frac{3.3}{\sqrt{451}} = 0.305](https://tex.z-dn.net/?f=M%20%3D%201.96%5Cfrac%7B3.3%7D%7B%5Csqrt%7B451%7D%7D%20%3D%200.305)
The lower end of the interval is the sample mean subtracted by M. So it is 7.74 - 0.305 = 7.435 hours
The upper end of the interval is the sample mean added to M. So it is 7.74 + 0.305 = 8.045 hours.
The 95% confidence interval to estimate μ is (7.435, 8.045) hours.
Question b:
90% confidence interval means that ![Z = 1.645](https://tex.z-dn.net/?f=Z%20%3D%201.645)
The margin of error is:
![M = 1.645\frac{21.96}{\sqrt{451}} = 1.701](https://tex.z-dn.net/?f=M%20%3D%201.645%5Cfrac%7B21.96%7D%7B%5Csqrt%7B451%7D%7D%20%3D%201.701)
43.78 - 1.701 = 41.079%
43.78 + 1.701 = 45.481%
The 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam is (41.079%, 45.481%).