Answer:
c) (26.295, 28.705)
Step-by-step explanation:
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{1-0.965}{2} = 0.0175[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.0175 = 0.9825[/tex], so [tex]z = 2.11[/tex]
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 = 2.11\frac{4}{\sqrt{49}} = 1.205[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 27.5 - 1.205 = 26.295 mi/gallon
The upper end of the interval is the sample mean added to M. So it is 27.5 + 1.205 = 28.705 mi/gallon
So the correct answer is:
c) (26.295, 28.705)
