Respuesta :
Answer:
C. (36337.32, 48968.68)
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{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find 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{9114}{\sqrt{8}} = 6315.68[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 42653 - 6315.68 = 36337.32.
The upper end of the interval is the sample mean added to M. So it is 42653 + 6315.68 = 48968.68.
So the correct answer is:
C. (36337.32, 48968.68)
The correct option is D.
[tex](35032.29,50273.71)[/tex]
Probability Sampling:
Probability sampling is described as a sampling method in which the person or researcher chooses samples from a larger population using a method based on the theory of probability. For the participant, it is necessary to choose a random selection.
Note that margin of Error [tex]E=\frac{t\alpha }{2}\ast \frac{s}{\sqrt{n}} \\[/tex]
Lower Bound [tex]X=\frac{-t\alpha }{2}\ast \frac{s}{\sqrt{n}} \\[/tex]
Upper Bound [tex]X=\frac{+t\alpha }{2}\ast \frac{s}{\sqrt{n}}[/tex]
Where,
[tex]\frac{\alpha }{2}=\frac{\left ( 1-confidence \ level \right )}{2}=0.025\\\frac{t\alpha }{2}=critical \ t \ for \ the \ confidence \ interval=2.364624252[/tex]
[tex]S[/tex]=sample standard deviation[tex]=9114[/tex]
[tex]n[/tex]=sample size[tex]=8[/tex]
[tex]df=n-1=7[/tex]
Thus, the Margin of Error[tex]E=7619.49468[/tex]
Lower bound[tex]=35033.50532[/tex]
Upper bound[tex]=50272.49468[/tex][
Thus, the confidence interval is[tex](35033.50532 \ , 50272.49468 )[/tex]
Learn more about the topic Probability Sampling: https://brainly.com/question/22241230
