Answer:
z-method: (66.26, 75.18)
Step-by-step explanation:
We have the standard deviation for the population, which means that we use the z-distribution to solve this question.
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{16.25}{\sqrt{51}} = 4.46[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 70.72 - 4.46 = 66.26
The upper end of the interval is the sample mean added to M. So it is 70.72 + 4.46 = 75.18
The correct answer is z-method: (66.26, 75.18)
