Question

A random sample of 8 Samsung Galaxy smartphones being sold over the Internet in 2013 had the following prices, in dollars:

249 349 299 249 149 135
199169
Assume the population standard deviation is alpha = 85.
If appropriate, construct a 99.5% confidence interval for the mean price for all phones of this type being sold on the Internet in 2013.

Answer 1

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

Explanation:

Sample mean:

Sum of all values divided by the number of values. So

`S_(m) = (249+349+299+249+149+135+199+169)/(8) = 224.75`

Confidence interval:

We have the standard deviation for the population, so the z-distribution is used.

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.995)/(2) = 0.0025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.0025 = 0.9975`, so Z = 2.807.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.807(85)/(√(8)) = 84.36`

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