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

Answer:

The 90% confidence interval for the population mean is $[0.5 - \frac{0.1645}{\sqrt{n}}, 0.5 + \frac{0.1645}{\sqrt{n}}]$, in which n is the number of students surveyed.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.9}{2} = 0.05$

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

$M = z\frac{\sigma}{\sqrt{n}}$

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

$M = 1.645\frac{0.1}{\sqrt{n}} = \frac{0.1645}{\sqrt{n}}$

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 $[0.5 - \frac{0.1645}{\sqrt{n}}, 0.5 + \frac{0.1645}{\sqrt{n}}]$, in which n is the number of students surveyed.