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
[tex]E=z\cdot \sigma/\sqrt{n}=1.96*0.7/\sqrt{26}=1.372/5.099=0.27[/tex]
Then, the upper and lower bounds are:
[tex]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[/tex]
Then, the 95% CI is
[tex]6.93\leq x\leq 7.47[/tex]
The margin of error for a 99% CI is
[tex]E=z\cdot \sigma/\sqrt{n}=2.58*0.7/\sqrt{26}=1.806/5.099=0.35[/tex]
Then, the upper and lower bounds are:
[tex]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[/tex]
Then, the 99% CI is
[tex]6.85\leq x\leq 7.55[/tex]