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.

We can estimate the margin of error with this formula:

And the margin of error is given by:

And we can rewrite the margin of error like this:

Where

For 95% of confidence the critical value is

The Standard error would be:

For 99% of confidence the critical value is

And the new margin of error would be:

And then the interval would be given by:

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".

represent the sample mean for the sample  

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:

  (1)

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

We can estimate the mean with this formula:

We can estimate the margin of error with this formula:

And the margin of error is given by:

And we can rewrite the margin of error like this:

Where

For 95% of confidence the critical value is

The Standard error would be:

For 99% of confidence the critical value is

And the new margin of error would be:

And then the interval would be given by:

because 1.2 isnt half of 6 duh

step-by-step explanation: