Respuesta :
Answer:
The 99.5% confidence interval for the mean price for all phones of this type being sold on the Internet in 2013 is between $140.39 and $309.11
Step-by-step explanation:
Sample mean:
Sum of all values divided by the number of values. So
[tex]S_{m} = \frac{249+349+299+249+149+135+199+169}{8} = 224.75[/tex]
Confidence interval:
We have the standard deviation for the population, so the z-distribution is used.
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 - 0.995}{2} = 0.0025[/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 - 0.0025 = 0.9975[/tex], so Z = 2.807.
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 = 2.807\frac{85}{\sqrt{8}} = 84.36[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 224.75 - 84.36 = $140.39
The upper end of the interval is the sample mean added to M. So it is 224.75 + 84.36 = $309.11
The 99.5% confidence interval for the mean price for all phones of this type being sold on the Internet in 2013 is between $140.39 and $309.11
