Question

A survey organization has used the methods of our class to construct an approximate 95% confidence interval for the mean annual income of households in a county. The interval runs from $66,000 to $70,000 . If possible, find an approximate 99% confidence interval for the mean annual income of households in the county. If this is not possible, explain why not.

Answer 1

Answer:

$\bar X = \frac{66000+70000}{2}= 68000$

We can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000$

And the margin of error is given by:

$ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

And we can rewrite the margin of error like this:

$ME =z_{\alpha/2}*SE$

Where $SE= \frac{\sigma}{\sqrt{n}}$

For 95% of confidence the critical value is $z_{\alpha/2}= \pm 1.96$

The Standard error would be:

$SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408$

For 99% of confidence the critical value is $z_{\alpha/2}= \pm 2.58$

And the new margin of error would be:

$ME = 2.58* 1020.408 = 2632.653$

And then the interval would be given by:

$Lower = 68000- 2632.653 = 65367.347$

$Upper = 68000+ 2632.653 = 70632.653$

Step-by-step explanation:

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The 95% confidence interval is given by (66000 , 70000)

We can estimate the mean with this formula:

$\bar X = \frac{66000+70000}{2}= 68000$

We can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000$

And the margin of error is given by:

$ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

And we can rewrite the margin of error like this:

$ME =z_{\alpha/2}*SE$

Where $SE= \frac{\sigma}{\sqrt{n}}$

For 95% of confidence the critical value is $z_{\alpha/2}= \pm 1.96$

The Standard error would be:

$SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408$

For 99% of confidence the critical value is $z_{\alpha/2}= \pm 2.58$

And the new margin of error would be:

$ME = 2.58* 1020.408 = 2632.653$

And then the interval would be given by:

$Lower = 68000- 2632.653 = 65367.347$

$Upper = 68000+ 2632.653 = 70632.653$