Problem 36 - Basic Statistics for Business and Economics with Student

Question

A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent): $$\begin{array}{|llllllll|}\hline 4.8 & 5.3 & 6.5 & 4.8 & 6.1 & 5.8 & 6.2 & 5.6 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude that the 30 -year mortgage rate for small banks is less than 6 percent? Estimate the \(p\) -value.

Step-by-Step Solution

Answer

We fail to reject the null hypothesis; the rate is not conclusively less than 6% at a 0.01 significance level.

Step 1: State the Hypotheses

First, we need to state the null and alternative hypotheses. The null hypothesis, \( H_0 \), is that the mean 30-year mortgage rate at small banks is 6 percent or more, \( \mu \geq 6 \). The alternative hypothesis, \( H_a \), is that the mean rate is less than 6 percent, \( \mu < 6 \).

Step 2: Collect the Data

Use the given sample data: \( 4.8, 5.3, 6.5, 4.8, 6.1, 5.8, 6.2, 5.6 \). Calculate the sample mean \( \bar{x} \) and sample standard deviation \( s \).

Step 3: Calculate the Sample Mean and Standard Deviation

Calculate the sample mean: \( \bar{x} = \frac{4.8 + 5.3 + 6.5 + 4.8 + 6.1 + 5.8 + 6.2 + 5.6}{8} = 5.6375 \). Calculate the sample standard deviation: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \approx 0.6551 \].

Step 4: Determine the Test Statistic

The test statistic for a one-sample t-test is given by \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]. For our data, \( \mu = 6 \), \( n = 8 \), so \[ t = \frac{5.6375 - 6}{0.6551/\sqrt{8}} \approx -1.512 \].

Step 5: Find the Critical Value and Decision Rule

With \( \alpha = 0.01 \) and \( n-1 = 7 \) degrees of freedom, find the critical t-value from the t-distribution table: approximately -2.998. If the calculated t-statistic is less than -2.998, reject the null hypothesis.

Step 6: Compare the Test Statistic to the Critical Value

The calculated t-statistic is -1.512. Since -1.512 is greater than -2.998, we fail to reject the null hypothesis.

Step 7: Estimate the p-value

Using a t-distribution table or calculator, find the p-value for \( t = -1.512 \) with 7 degrees of freedom. The p-value is approximately 0.085, which is greater than 0.01.

Key Concepts

Statistical Significance

In hypothesis testing, statistical significance helps us determine whether a result is likely due to chance or if there is enough evidence to support a given hypothesis.
In our example, we are trying to find out if the average 30-year mortgage rate is genuinely less than 6 percent.
The significance level, denoted by \( \alpha \), quantifies the threshold of risk we are willing to take. Here, it is set to 0.01, meaning we accept a 1% chance of incorrectly rejecting the null hypothesis.
Statistically significant results have a low \( p \)-value, less than our chosen \( \alpha \), leading us to reject the null hypothesis.

t-test

A t-test is a statistical test used to compare the sample mean to a known value, often determining if the difference is significant.
In our mortgage rate exercise, we apply a one-sample t-test to evaluate whether the sample mean mortgage rate is different from 6 percent.
We calculate the t-statistic using the formula:

• \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)

where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean (6%), \( s \) is the sample standard deviation, and \( n \) is the sample size.

p-value

The \( p \)-value measures the probability of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true.
In our situation, a \( p \)-value of 0.085 indicates that there is an 8.5% chance of seeing such an extreme sample mean, assuming the true mean is 6 percent.
Since this \( p \)-value is higher than our significance level (0.01), we cannot reject the null hypothesis.
Calculating the \( p \)-value helps decide how far the observed data deviates from the expected under the null hypothesis.

Sample Mean

The sample mean represents the average value of a set of data points.
In our study, we calculate the sample mean of mortgage rates from eight banks, which came out to be 5.6375 percent.
The sample mean provides a central point of comparison for evaluating the differences from a hypothesized population mean.
It plays a critical role in conducting a t-test and helps determine whether observed data provide enough evidence to support the alternative hypothesis.

Standard Deviation

Standard deviation is a measure of variability or dispersion within a set of data points.
In this exercise, the standard deviation evaluates how spread out the individual mortgage rates are from the sample mean of 5.6375 percent.
A calculated standard deviation of approximately 0.6551 helps provide context in the t-test formula, correcting for variability in the sample.
Understanding standard deviation is essential for identifying how consistent or varied the dataset is and for assessing statistical significance during hypothesis testing.