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](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20%5Csigma%2F%5Csqrt%7Bn%7D%3D1.96%2A0.7%2F%5Csqrt%7B26%7D%3D1.372%2F5.099%3D0.27)
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](https://tex.z-dn.net/?f=LL%3D%5Cbar%20x-z%5Ccdot%5Csigma%2F%5Csqrt%7Bn%7D%3D7.2-0.27%3D6.93%5C%5C%5C%5C%20UL%3D%5Cbar%20x%2Bz%5Ccdot%5Csigma%2F%5Csqrt%7Bn%7D%20%3D7.2%2B0.27%3D7.47)
Then, the 95% CI is
![6.93\leq x\leq 7.47](https://tex.z-dn.net/?f=6.93%5Cleq%20x%5Cleq%207.47)
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](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20%5Csigma%2F%5Csqrt%7Bn%7D%3D2.58%2A0.7%2F%5Csqrt%7B26%7D%3D1.806%2F5.099%3D0.35)
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](https://tex.z-dn.net/?f=LL%3D%5Cbar%20x-z%5Ccdot%5Csigma%2F%5Csqrt%7Bn%7D%3D7.2-0.35%3D6.85%5C%5C%5C%5C%20UL%3D%5Cbar%20x%2Bz%5Ccdot%5Csigma%2F%5Csqrt%7Bn%7D%20%3D7.2%2B0.35%3D7.55)
Then, the 99% CI is
![6.85\leq x\leq 7.55](https://tex.z-dn.net/?f=6.85%5Cleq%20x%5Cleq%207.55)
Answer:
B.
Step-by-step explanation:
You have to find the total of tickets which is 85 and find the non-winning tickets which is 63. So, the answer is 63:85.
Answer:the answer is
Step-by-step explanation:c