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

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.

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yotaniwc
Step 1
Given that,
Sample size $$\displaystyle{n}={25}$$
Sample mean $$\displaystyle\overline{{{x}}}={0.0695}$$
Population standard deviation $$\displaystyle{s}={0.006}$$
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:
$$\displaystyle{C}{I}=\overline{{{x}}}\pm{z}_{{{c}}}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={0.0695}\pm{1.645}{\left({\frac{{{0.006}}}{{\sqrt{{{25}}}}}}\right)}$$
$$\displaystyle={0.0695}\pm{0.00197}$$
$$\displaystyle={\left({0.068},\ {0.071}\right)}$$
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:
$$\displaystyle{C}{I}=\overline{{{x}}}\pm{z}_{{{c}}}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$
$$\displaystyle={0.0695}\pm{2.58}{\left({\frac{{{0.006}}}{{\sqrt{{{25}}}}}}\right)}$$
$$\displaystyle={0.0695}\pm{0.0031}$$
$$\displaystyle={\left({0.066},\ {0.073}\right)}$$
The 99% confidence intervals for the population mean 30-year fixed mortgage rate is from 6.6% to 7.3%
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lovagwb
Step 1
Given that:
$$\displaystyle{n}={25}$$S
$$\displaystyle\overline{{{x}}}={6.95}$$
$$\displaystyle\sigma={0.6}$$
To find C.I we have formula
$$\displaystyle\overline{{{x}}}\pm{Z}_{{{\frac{{\alpha}}{{{2}}}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle{a}+\alpha={0.10}$$
$$\displaystyle{l}\times{0}\times{S}$$
$$\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={1.645}$$
From equal 1
$$\displaystyle{6.95}\pm{\left({1.645}\right)}{\frac{{{0.6}}}{{\sqrt{{{25}}}}}}$$
$$\displaystyle{6.95}\pm{0.1974}$$
$$\displaystyle{\left({6.7526},\ {7.1474}\right)}$$
$$\displaystyle\therefore{90}\%\ {C}.{I}{\left({6.75},\ {7.15}\right)}$$
at $$\displaystyle\alpha={0.01}$$
$$\displaystyle{l}\times{0}\times{s}$$
$$\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}={2.576}$$
From equal 1
$$\displaystyle{6.95}\pm{\left({2.576}\right)}{\left[{\frac{{{0.6}}}{{\sqrt{{{25}}}}}}\right]}$$
$$\displaystyle{6.95}\pm{0.3091}$$
$$\displaystyle{\left({6.6409},\ {7.2591}\right)}$$
$$\displaystyle\therefore{99}\%\ {C}.{I}{\left({6.64},\ {7.26}\right)}$$
karton
Step 1
$\begin{array}{|c|c|}\hline \text{sample mean}\ \bar{x} & 6.950 \\ \hline \text{sample size}\ n= & 25.00 \\ \hline \text{std deviation}\ \sigma= & 0.600 \\ \hline \text{std error}\ =\sigma x=\frac{\sigma}{\sqrt{n}}= & 0.1200 \\ \hline \end{array}$
For 90% level:
$\begin{array}{|c|c|}\hline \text{for 90% Cl value of}\ z= & 1.645 \\ \hline \text{margin of error}\ E=z\times std\ error & 0.20 \\ \hline \text{lower bound=sample mean-E=} & 6.75 \\ \hline \text{Upper bound=sample mean+E=} & 7.15 \\ \hline \end{array}$
For 99% level:
$\begin{array}{|c|c|}\hline \text{for 99% Cl value of}\ z= & 2.576 \\ \hline \text{margin of error}\ E=z\times std\ error & 0.31 \\ \hline \text{lower bound=sample mean-E=} & 6.64 \\ \hline \text{Upper bound=sample mean+E=} & 7.26 \\ \hline \end{array}$
$\begin{array}{|c|c|}\hline \text{Confidence level} & \text{confidence interval} \\ \hline 90\% & 6.75\%\ to\ 7.15\% \\ \hline 99\% & 6.64\%\ to\ 7.26\% \\ \hline \end{array}$