Question

A MORTGAGE SPECIALIST WOULD LIKE TO ANALYZE THE AVERAGE RATES FOR ATLANTA, GEORGIA. HE COLLECTS DATA ON THE ANNUAL PERCENTAGE RATES (APR IN%) FOR 30 YEAR FIXED LOANS AS SHOWN IN THE FOLLOWING TABLE. IF HE IS WILLING TO ASSUME THAT THESE RATES ARE RANDOMLY DRAWN FROM A NORMALLY DISTRIBUTED POPULATION, CAN HE CONCLUDE THAT THE MEAN MORTGAGE RATE FOR THE POPULATION EXCEEDS 4.10? TEST THE HYPOTHESIS AT A 5% LEVEL OF SIGNIFICANCE.

financial institution apr

g squared financial 4.165

best possible mortgage 4.805

hersch financial group 4.400

total mortgages services 4.740

wells fargo 4.290

quicken loans 4.170

amerisave 4.705

CALCULATE THE VALUE OF TEST STATISTICS. ( ROUND IMMEDIATE CALCULATIONS TO AT LEAST 4 DECIMALS AND FINAL ANSWER TO 2 DECIMALS)

Answer 1

Let $\small \mu$ be the true average mortgage rate for Atlanta, Georgia (the population).

The mortgage specialist wants to test if the true average mortgage rate for this population, exceeds 4.10

That is we want to test the following hypotheses

$\small \begin{align*} &H_0:\mu=4.10\leftarrow\text{null hypothesis: The average mortgage rate is 4.10\%}\\ &H_a:\mu>4.10\leftarrow\text{alternative hypothesis: The average mortgage rate exceeds 4.10\%}\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

This is a right tailed (one tailed) test (The alternative hypothesis has ">")

We do not know the population standard deviation of APR and the sample size n=7 is less than 30. But the rates are normally distributed. Hence we can do a small sample analysis. We will use t-distribution for the sample statistics.

We have the following sample information

n=7 is the sample size

$\small \begin{align*} \bar{x}=\frac{\sum_i x_i}{n}=\frac{4.165+ 4.805+ 4.4+ 4.74+ 4.29+ 4.17+ 4.705}{7}=4.468 \end{align*}$ is the sample mean

$\small \begin{align*} s&=\sqrt\frac{\sum_i( x_i-\bar{x})^2}{n-1}\\ &=\sqrt\frac{(4.165-4.468)^2+ (4.805-4.468)^2+ (4.4-4.468)^2+ (4.74-4.468)^2+ (4.29-4.468)^2+ (4.17-4.468)^2+ (4.705-4.468)^2}{7-1}\\&=0.2771 \end{align*}$is the sample standard deviation.

Since we do not know the population standard deviation, we will use the sample to estimate it

$\small \begin{align*} \hat{\sigma}=s=0.2771 \end{align*}$

The standard error of mean is

$\small \begin{align*} \hat{\sigma}_{\bar{x}}=\frac{\hat{\sigma}}{\sqrt{n}}=\frac{0.2771}{\sqrt{7}}=0.1047 \end{align*}$

The hypothesized value of true mean (from the null hypothesis) is $\small \begin{align*} \mu_{H_0}=4.10 \end{align*}$

The test statistics is

$\small \begin{align*} t=\frac{\bar{x}-\mu_{H_0}}{\hat{\sigma}_{\bar{x}}}=\frac{4.468-4.10 }{0.1047}=3.51 \end{align*}$

ans: The value of test statistics is 3.51

The degrees of freedom of the t-statistics is n-1=7-1=6

this is a right tail test. Hence we look only at the area under the right tail. The critical value of t ($\small \begin{align*} P(T>t_{\alpha})=0.05 \end{align*}$) is obtained from t tables for level of significance, alpha=0.05, and df=6. We get a value of 1.943. that is P(T>1.943) = 0.05.

Hence the value of critical value is 1.943

We will reject the null hypothesis if the test statistics is greater than the critical value.

Here, the test statistics of 3.51 is greater than the critical value 1.943. Hence we reject the null hypothesis.

We conclude that there is sufficient evidence to support that claim that the mean mortgage rate for the population exceeds 4.10.