Respuesta :
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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96\frac{3.3}{\sqrt{451}} = 0.305[/tex]
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 [tex]Z = 1.645[/tex]
The margin of error is:
[tex]M = 1.645\frac{21.96}{\sqrt{451}} = 1.701[/tex]
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%).
