Question

One measure of the state of the economy is the amount of money homeowners pay on their mortgage each month. To determine the extent of change between this year and 5 years ago, a random sample of 150 homeowners was drawn. The monthly mortgage payments for each homeowner for this year and for 5 years ago was recorded (the amounts were adjusted so that we could compare constant dollars). One would like to use the data to estimate the mean difference in monthly mortgage payments with 90% confidence.

Which formula is appropriate for the scenario described?

Using the information below, calculate the confidence interval, rounding your limits to 3 decimal places.

Mean Std. Error Critical Value 2016-2011 -109.456 16.196 1.655

Lower limit: _

Upper limit: _

Interpret: With 90% confidence, we estimate the proportion of mortgage payments/mean difference/mean mortgage payment in 2016/2011 is between _ and _ more than that of 2016/2011  .

Answer 1

The confidence interval is given by $\bar X \pm t * SE$

The critical value is given as 1.655 for 90% confidence.

We have $\bar{X} = -109.456$ , t = 1.655 and SE = 16.196

Thus the confidence interval is

$=-109.456 \pm 1.655* 16.196$

$= -109.456 \pm 26.804$

Thus lower limit = -109.456 - 26.804 = -136.26

Upper limit = -109.456 + 26.804 = -82.646

With 90% confidence, the confidence interval for the mean difference in monthly payments in 2011 as compared to 2006 is -136.26 and -82.646

That is $-136.26 \leq \mu \leq -82.646$