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Chapter 8: Problem 160

A fire insurance company thought that the mean distance from a home to the nearest fire department in a suburb of Chicago was at least 4.7 miles. It set its fire insurance rates accordingly. Members of the community set out to show that the mean distance was less than 4.7 miles. This, they thought, would convince the insurance company to lower its rates. They randomly identified 64 homes and measured the distance to the nearest fire department from each. The resulting sample mean was \(4.4 .\) If \(\sigma=2.4\) miles, does the sample show sufficient evidence to support the community's claim at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
Yes, there is sufficient evidence to support the community's claim at the 0.05 level of significance that the mean distance from a home to the nearest fire department is less than 4.7 miles.

Step by step solution

01

Set up the Hypotheses

The null hypothesis (H0) is that the mean distance is equal to or greater than 4.7 miles, while the alternative hypothesis (H1) is that the mean distance is less than 4.7 miles. Mathematically, this can be represented as: H0: \( \mu \geq 4.7 \), H1: \( \mu < 4.7 \)
02

Calculate the Test Statistic

Learning that the population standard deviation (\( \sigma \)) is 2.4, the number of homes (n) is 64 and the sample mean (\( \bar{x} \)) is 4.4, the test statistic (z) can be calculated using the formula: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Substituting the given values, we get \( z = \frac{4.4 - 4.7}{\frac{2.4}{\sqrt{64}}} = -2 \)
03

Find the Critical Value and Make a Decision

The critical value for a one-tailed test at the 0.05 level of significance is -1.645. Comparing the computed test statistic with the critical value, we have -2 < -1.645. Hence, we reject the null hypothesis. There's sufficient evidence at the 0.05 level of significance to support the community's claim that the mean distance to a fire department from a home is less than 4.7 miles.

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Key Concepts

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

Null Hypothesis
The null hypothesis is the default statement that there is no effect or no difference. It is what we assume to be true before we collect data. In the context of the insurance company case study, the null hypothesis posits that the mean distance from homes to the nearest fire department is at least 4.7 miles. Mathematically expressed as:
H0: \( \boldsymbol{\mu} \geq 4.7 \).
Understanding the null hypothesis is critical as it forms the basis for statistical testing. The objective of hypothesis testing is to determine whether there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
Alternative Hypothesis
Conversely, the alternative hypothesis represents a statement that indicates a difference or effect that contradicts the null hypothesis. For the community challenging the insurance company's rates, the alternative hypothesis claims that the mean distance is less than 4.7 miles, suggesting the insurance rates may be unjustly high. This is symbolized as:
H1: \( \boldsymbol{\mu} < 4.7 \).
In hypothesis testing, the alternative hypothesis is what the study aims to support. This is achieved by providing evidence strong enough to reject the null hypothesis.
Test Statistic
The test statistic is a crucial component calculated from the sample data that helps us make a decision about the hypothesis. It reflects how much the observed sample statistic differs from the parameter specified in the null hypothesis, measured in units of standard error. Here, the test statistic is a z-value found using:
\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \].
For our example, a test statistic of -2 indicates that the sample mean is 2 standard errors below the hypothesized population mean of 4.7 miles. The more negative the test statistic, the stronger the evidence against the null hypothesis.
Level of Significance
The level of significance, denoted as alpha (\( \alpha \)), is the probability of rejecting the null hypothesis when it is actually true. It essentially acts as a threshold for how much chance we're willing to accept. The lower the \( \alpha \), the less likely we are to make a type I error鈥攔ejecting the null hypothesis erroneously. In the insurance company scenario, \( \alpha = 0.05 \) means there is a 5% risk of concluding that the mean distance is less than 4.7 miles when it is not.
Population Standard Deviation
The population standard deviation (\( \sigma \)) is a measure of the variability or spread of a set of data. Knowing the population standard deviation is significant as it helps in calculating the test statistic and making inferences about the population. In this example, \( \sigma = 2.4 \) miles implies that the distances from homes to fire departments are dispersed around the mean distance by an average of 2.4 miles.
Sample Mean
The sample mean (\( \bar{x} \)) is the average distance calculated from the randomly selected sample of homes. It's the estimated value of the true population mean that we use in hypothesis testing. The sample mean for our exercise is 4.4 miles, and it is this value against which the null hypothesis is tested. The sample mean serves as the basis for calculating the test statistic and ultimately determining whether there is sufficient evidence to support the alternative hypothesis.
Critical Value
The critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. If the test statistic falls beyond the critical value, the null hypothesis is rejected. Critical values are determined by the level of significance (\( \alpha \)) and the nature of the test. In the problem at hand, the critical value for a one-tailed z-test at the 0.05 significance level is -1.645. Since our test statistic of -2 is less than -1.645, we can reject the null hypothesis, thus supporting the community's claim.

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Most popular questions from this chapter

Professor Hart does not believe the statement "The mean distance commuted daily by the nonresident students at our college is no more than 9 miles." State the null and alternative hypotheses he would use to challenge this statement.

The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12 minutes and a standard deviation of 2.3 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 28 customers visiting her store this week resulted in a mean of 10.9 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12 minutes? Use \(\alpha=0.02\)

A random sample of the scores of 100 applicants for clerk-typist positions at a large insurance company showed a mean score of \(72.6 .\) The preparer of the test maintained that qualified applicants should average \(75.0 .\) a. Determine the \(99 \%\) confidence interval for the mean score of all applicants at the insurance company. Assume that the standard deviation of test scores is 10.5 b. Can the insurance company conclude that it is getting qualified applicants (as measured by this test)?

a. What decision is reached when the \(p\) -value is greater than \(\alpha ?\) b. What decision is reached when \(\alpha\) is greater than the \(p\) -value?

When a parachute is inspected, the inspector is looking for anything that might indicate the parachute might not open. a. State the null and alternative hypotheses. b. Describe the four possible outcomes that can result depending on the truth of the null hypothesis and the decision reached. c. Describe the seriousness of the two possible errors.
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