Question

In order to estimate the mean 30-year fixed mortgage rate for a home loan in the United States, a random sample of 26 recent loans is taken. The average calculated from this sample is 7.20%. It can be assumed that 30-year fixed mortgage rates are normally distributed with a standard deviation of 0.7%. Compute 95% and 99% confidence intervals for the population mean 30-year fixed mortgage rate.

Answer 1

Answer:

The 95% CI is (6.93% , 7.47%)

The 99% CI is (6.85% , 7.55%)

Step-by-step explanation:

We have to estimate two confidence intervals (95% and 99%) for the population mean 30-year fixed mortgage rate.

We know that the population standard deviation is 0.7%.

The sample mean is 7.2%. The sample size is n=26.

The z-score for a 95% CI is z=1.96 and for a 99% CI is z=2.58.

The margin of error for a 95% CI is

$E=z\cdot \sigma/\sqrt{n}=1.96*0.7/\sqrt{26}=1.372/5.099=0.27$

Then, the upper and lower bounds are:

$LL=\bar x-z\cdot\sigma/\sqrt{n}=7.2-0.27=6.93\\\\ UL=\bar x+z\cdot\sigma/\sqrt{n} =7.2+0.27=7.47$

Then, the 95% CI is

$6.93\leq x\leq 7.47$

The margin of error for a 99% CI is

$E=z\cdot \sigma/\sqrt{n}=2.58*0.7/\sqrt{26}=1.806/5.099=0.35$

Then, the upper and lower bounds are:

$LL=\bar x-z\cdot\sigma/\sqrt{n}=7.2-0.35=6.85\\\\ UL=\bar x+z\cdot\sigma/\sqrt{n} =7.2+0.35=7.55$

Then, the 99% CI is

$6.85\leq x\leq 7.55$