Chapter 8.2, Problem 32E

Mortgage rates: Following are interest rates (annual percentage rates) for a 30-year fixed rate mortgage from a sample of lenders in Macon. Georgia on June 20. 2013. It is reasonable to assume that the population is approximately normal.

1. Construct a 99 confidence interval for the mean rate.
2. One week earlier. The mean rate was 4.050%. A mortgage broker claims that the mean rate is now higher. Based on the confidence interest. Is this a reasonable claim? Explain.

To determine

The $99\%$ confidence interval for the mean rate.

Answer to Problem 32E

From the above MINITAB output, the $99\%$ confidence interval for the mean rate is $\left(4.1190,4.5248\right)$. That is, $4.1190<\mu <4.5248$.

Explanation of Solution

Given information:

Following are interest rates (annual percentage rates) for a $30-$ year fixed rate mortgage from a sample of lenders in Macon, Georgia on June $20,2013$.

$\begin{array}{l}4.750\\ 4.125\end{array}$ $\begin{array}{l}4.750\\ 4.125\\ \\ 4.375\\ 4.250\\ \\ 4.176\\ 3.950\\ \\ 4.679\\ 4.191\\ \\ 4.426\\ 4.299\\ \\ \end{array}$

$\begin{array}{l}4.375\\ 4.250\end{array}$

$\begin{array}{l}4.176\\ 3.950\end{array}$

$\begin{array}{l}4.679\\ 4.191\end{array}$

$\begin{array}{l}4.426\\ 4.299\end{array}$

$\begin{array}{l}4.227\\ 4.415\end{array}$

Population is normally distributed.

Calculation:

MINITAB is used to construct $99\%$ confidence interval for the mean rate.

MINITAB procedure:

Step $1:$ Choose Stat $>$ Basic Statistics $>1$ -Sample $t$.

Step $2:$ In Summarized data, enter the sample size $12$ and mean $4.32$.

Step $3:$ In Standard deviation, enter a value $0.226$.

Step $4:$ Check Options, enter Confidence level as $99.0$.

Step $5:$ Choose not equal in alternative.

Step $6:$ Click OK in all dialogue boxes.

MINITAB output:

One-Sample $\text{T}$

$\begin{array}{l}\text{N Mean StDev }\text{99\% CI}\\ \text{12 4}\text{.32 0}\text{.226 }\left(4.15,\text{ 4}\text{.49}\right)\end{array}$

From the above MINITAB output, the $99\%$ confidence interval for the true mean value of all mortgage rate is $(4.15,4.49)$.

To determine

Whether claim made in the problem statement is contradictory to the confidence interval.

Answer to Problem 32E

No, it is a reasonable claim.

Explanation of Solution

Given information:

Following are interest rates (annual percentage rates) for a $30-$ year fixed rate mortgage from a sample of lenders in Macon, Georgia on June $20,2013$.

$\begin{array}{l}4.750\\ 4.125\end{array}$ $\begin{array}{l}4.750\\ 4.125\\ \\ 4.375\\ 4.250\\ \\ 4.176\\ 3.950\\ \\ 4.679\\ 4.191\\ \\ 4.426\\ 4.299\\ \\ \end{array}$

$\begin{array}{l}4.375\\ 4.250\end{array}$

$\begin{array}{l}4.176\\ 3.950\end{array}$

$\begin{array}{l}4.679\\ 4.191\end{array}$

$\begin{array}{l}4.426\\ 4.299\end{array}$

$\begin{array}{l}4.227\\ 4.415\end{array}$

Population is normally distributed.

One week earlier the mean rate was $4.050\%$.

A claim is made by broker that the mean rate is now higher.

Calculation:

From part $a$ of this exercise, the $99\%$ confidence interval for the true mean value of all mortgage rate is $(4.15,4.49)$. This mean there is a $99\%$ chance that the mean mortgage rate lies in the interval $(4.15,4.49)$ percentage.

Since this interval does contain the value $4.050$ % which is lower than the $99\%$ interval range. Therefore, mean rate is now lower. So claim made by broker is not reasonable.