Question

The cost of Internet access. How much do users pay for Internet service? Here are the monthly fees (in dollars) paid by a random sample of 50 users of commercial Internet service providers in August 2000: 11 Give a 95% confidence interval for the mean monthly cost of Internet access in August 2000.

Answer 1

`20.9-2.01(7.65)/(√(50))=18.73`

`20.9+2.01(7.65)/(√(50))=23.07`

The 95% confidence interval would be given by (18.73;23.07)

Explanation:

Assuming these data

20 40 22 22 21 21 20 10 20 20

20 13 18 50 20 18 15 8 22 25

22 10 20 22 22 21 15 23 30 12

9 20 40 22 29 19 15 20 20 20

20 15 19 21 14 22 21 35 20 22

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

First we need to find the sample mean with the following formula:

`\bar X= (\sum_(i=1)^n X_i)/(50)=20.9`

And in order to find the sample standard deviation we can use the following formula:

`s= \sqrt{(\sum_(i=1)^n (x_i -\bar x)^2)/(n-1)}=7.65`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=50-1=49`

Since the Confidence is 0.95or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that
`t_(\alpha/2)=2.01`

Now we have everything in order to replace into formula (1):

`20.9-2.01(7.65)/(√(50))=18.73`

`20.9+2.01(7.65)/(√(50))=23.07`

So on this case the 95% confidence interval would be given by (18.73;23.07)