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Chapter 10: Problem 30

A recent article in The Wall Street Journal reported that the home equity loan rate is now less than \(4 \% .\) A sample of eight small banks in the Midwest revealed the following home equity loan rates (in percent): \(\begin{array}{lllllll}3.6 & 4.1 & 5.3 & 3.6 & 4.9 & 4.6 & 5.0 & 4.4\end{array}\) At the .01 significance level, can we conclude that the home equity loan rate for small banks is less than \(4 \%\) ? Estimate the \(p\) -value.

Short Answer

Expert verified
Do not reject the null hypothesis; the average rate is not significantly less than 4%. The p-value is greater than 0.01.

Step by step solution

01

Formulate the Hypotheses

We need to establish the null and alternative hypotheses. The null hypothesis is that the average home equity loan rate is \( \mu \geq 4\% \). The alternative hypothesis is \( \mu < 4\% \). This is a left-tailed test.
02

Calculate the Sample Mean

First, calculate the mean of the sample data provided: \( \{3.6, 4.1, 5.3, 3.6, 4.9, 4.6, 5.0, 4.4\} \). The sample mean \( \bar{x} \) is computed as:\[ \bar{x} = \frac{3.6 + 4.1 + 5.3 + 3.6 + 4.9 + 4.6 + 5.0 + 4.4}{8} = 4.44 \]
03

Calculate the Sample Standard Deviation

Calculate the sample standard deviation \( s \). First, find the squared deviations from the mean and their sum:\[\sum_{i=1}^{n} (x_i - \bar{x})^2 = (3.6-4.44)^2 + (4.1-4.44)^2 + \ldots + (4.4-4.44)^2 = 4.852\]\[ s = \sqrt{\frac{4.852}{8-1}} = 0.74 \]
04

Compute the Test Statistic

Using the sample mean and standard deviation, calculate the test statistic:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{4.44 - 4}{0.74/\sqrt{8}} = 1.65 \]
05

Determine the Critical Value and Decision Rule

Find the critical value for \( t \) at the 0.01 significance level with \( n-1 = 7 \) degrees of freedom. Using the t-distribution table, the critical value for a left-tailed test at \( \alpha = 0.01 \) is approximately \(-2.998\). Our decision rule is: Reject \( H_0 \) if \( t < -2.998 \).
06

Compare Test Statistic and Critical Value

The calculated test statistic \( t = 1.65 \) is greater than the critical value \(-2.998\). Therefore, we do not reject the null hypothesis.
07

Estimate the P-value

Using a t-distribution table or calculator for 7 degrees of freedom, find the p-value corresponding to \( t = 1.65 \). The p-value is significantly greater than 0.01, indicating weak evidence against the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-test
A t-test is a fundamental statistical method used to determine if there is a significant difference between the mean of a sample and a known value or between the means of two samples. It helps us understand whether an observed effect is genuine or if it could have happened by random chance. In practice:
  • We use a t-test when the sample size is relatively small, usually less than 30.
  • The data should approximately follow a normal distribution.
  • It's commonly used in various types of hypothesis testing.
For our example, we're dealing with a one-sample t-test comparing the mean home equity loan rate of eight banks to see if it is indeed less than 4%. Our calculated t-statistic is 1.65, which tells us how far our sample mean is from the hypothesized population mean, in units of standard error.
significance level
The significance level, denoted as \( \alpha \), is a threshold set by the researcher before conducting a hypothesis test. It reflects the probability of rejecting the null hypothesis when it is actually true. Essentially, it is the risk we are willing to take for a false positive:
  • Common choices for \( \alpha \) are 0.05, 0.01, and 0.10.
  • A lower \( \alpha \) (like 0.01) indicates stricter criteria for significance.
  • In hypothesis testing, we compare the p-value with our chosen \( \alpha \) to decide whether to reject the null hypothesis.
In our problem, we chose a significance level of 0.01, which means we want to be 99% confident in our decision regarding the bank's loan rates. Because our test statistic doesn鈥檛 surpass the critical value associated with this \( \alpha \), we lack sufficient evidence to reject the null hypothesis.
p-value
The p-value is a measure that helps us understand the strength of the results from a statistical hypothesis test. It tells us the probability of obtaining results that are as extreme as the observed results, assuming the null hypothesis is true. Here's how it works:
  • If the p-value is less than \( \alpha \), we reject the null hypothesis.
  • The smaller the p-value, the stronger the evidence against the null hypothesis.
  • A large p-value suggests that the observed data is consistent with the null hypothesis.
In our exercise, the p-value for our t-statistic of 1.65 is larger than 0.01, indicating there's not enough evidence to say the bank rates are less than 4%. Hence, the null hypothesis remains unchallenged.
null hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement asserting that there is no effect or no difference. It serves as a starting point for statistical analysis. In hypothesis testing:
  • It is usually phrased in a way to represent the status quo or a default position.
  • The goal is to test the null hypothesis against an alternative hypothesis.
  • Evidence must be statistically significant to reject the null hypothesis.
In our scenario, the null hypothesis states that the home equity loan rate is \( \mu \geq 4\% \). After conducting the t-test, we found that the evidence was not substantial enough to reject this hypothesis, aligning with the World Journal's claim that the rate is not significantly less than 4% at the 0.01 significance level.

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Most popular questions from this chapter

For Exercises 5-8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The manufacturer of the \(X-15\) steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of 15 agents are: \(\begin{array}{lllllllllll}53 & 57 & 50 & 55 & 58 & 54 & 60 & 52 & 59 & 62 & 60 & 60 & 51 & 59 & 56\end{array}\) At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Estimate the \(p\) -value.

Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is 21 , and the population standard deviation is 5 . Conduct the following test of hypothesis using the .05 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 20 \\ H_{1}: \mu>20 \end{array} $$

According to the Census Bureau, 3.13 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.20 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.13 persons?

The management of White Industries is considering a new method of assembling its golf cart. The present method requires a mean time of 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?
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