Question

Students at Elon University were surveyed and the average commute distance to campus is 0.5 mile. Assume the population standard deviation is 0.1 mile. Please construct a 90% confidence interval for the population mean.

a. [0.472, 0.528]
b. [0.479, 0.521)
c. [0.463, 0.534]
d. [0.476, 0.524]

Answer 1

The 90% confidence interval for the population mean is
`[0.5 - (0.1645)/(√(n)), 0.5 + (0.1645)/(√(n))]`, in which n is the number of students surveyed.

Explanation:

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

`\alpha = (1 - 0.9)/(2) = 0.05`

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 - 0.05 = 0.95`, so Z = 1.645.

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 = 1.645(0.1)/(√(n)) = (0.1645)/(√(n))`

The lower end of the interval is the sample mean subtracted by M, while the upper end is M added to the sample mean of 0.5. Thus, the confidence interval is of:

The 90% confidence interval for the population mean is
`[0.5 - (0.1645)/(√(n)), 0.5 + (0.1645)/(√(n))]`, in which n is the number of students surveyed.