In order to estimate the mean 30-year fixed mortgage rate fo

Marenonigt

Marenonigt

Answered question

2021-12-26

In order to estimate the mean 30-year fixed mortgage rate for a home loan in the United States, a random sample of 25 recent loans is taken. The average calculated from this sample is 6.95%. It can be assumed that 30-year fixed mortgage rates are normally distributed with a standard deviation of 0.6%. Compute 90% and 99% confidence intervals for the population mean 30-year fixed mortgage rate.

Answer & Explanation

yotaniwc

yotaniwc

Beginner2021-12-27Added 34 answers

Step 1
Given that,
Sample size n=25
Sample mean x=0.0695
Population standard deviation s=0.006
Step 2
90% confidence intervals for the population mean 30-year fixed mortgage rate:
Critical value: The two tailed z critical value at 90% confidence level is 1.645
Calculation: The 90% confidence intervals for the population mean 30-year fixed mortgage rate can be calculated as follows:
CI=x±zc(σn)
=0.0695±1.645(0.00625)
=0.0695±0.00197
=(0.068, 0.071)
The 90% confidence intervals for the population mean 30-year fixed mortgage rate is from 6.8% to 7.1%.
Step 3
99% confidence intervals for the population mean 30-year fixed mortgage rate:
Critical value: The two tailed z critical value at 99% confidence level is 2.58.
Calculation: The 99% confidence intervals for the population mean 30-year fixed mortgage rate can be calculated as follows:
CI=x±zc(σn)
=0.0695±2.58(0.00625)
=0.0695±0.0031
=(0.066, 0.073)
The 99% confidence intervals for the population mean 30-year fixed mortgage rate is from 6.6% to 7.3%
lovagwb

lovagwb

Beginner2021-12-28Added 50 answers

We have that: 
n=25
x=6.95 
σ=0.6 
To determine C.I we have formula 
x±Zα2σn 
a+α=0.10 
l×0×S 
Zα2=1.645 
using equal 1
6.95±(1.645)0.625 
6.95±0.1974 
(6.7526, 7.1474) 
90% C.I(6.75, 7.15) 
at α=0.01 
l×0×s 
Zα2=2.576 
using equal 1
6.95±(2.576)[0.625] 
6.95±0.3091 
(6.6409, 7.2591) 
99% C.I(6.64, 7.26)

karton

karton

Expert2022-01-04Added 613 answers

 sample mean x¯6.950sample size n=25.00std deviation σ=0.600std error =σx=σn=0.1200 For 90% level: for 90% Cl value of z=1.645margin of error E=z×std error0.20lower bound=sample mean-E=6.75Upper bound=sample mean+E=7.15 For 99% level: for 99% Cl value of z=2.576margin of error E=z×std error0.31lower bound=sample mean-E=6.64Upper bound=sample mean+E=7.26 Confidence levelconfidence interval90%6.75% to 7.15%99%6.64% to 7.26%

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