Answer:
a) The 95% confidence interval for the average GPA of all of UCLA's entering freshmen is (4.11, 4.31).
b) 4.05 is not part of the confidence interval, which means that it does not support this claim.
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 Z-table as such z has a p-value 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{0.75}{\sqrt{250}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 4.21 - 0.1 = 4.11
The upper end of the interval is the sample mean added to M. So it is 4.21 + 0.1 = 4.31
The 95% confidence interval for the average GPA of all of UCLA's entering freshmen is (4.11, 4.31).
b. UCLA claims the average GPA for entering freshmen is 4.05. Does your confidence interval support this claim?
4.05 is not part of the confidence interval, which means that it does not support this claim.
