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An economist wishes to determine whether people are driving less than in the past. In one region of the country the number of miles driven per household per year in the past was 18.59 thousand miles. A sample of 15 households produced a sample mean of 16.23 thousand miles for the last year, with sample standard deviation 4.06 thousand miles. Assuming a normal distribution of household driving distances per year, perform the relevant test at the \(5 \%\) level of significance.

Short Answer

Expert verified
Reject the null hypothesis; people are driving less.

Step by step solution

01

Define the null and alternative hypotheses

Let's define the hypotheses for this test. - Null Hypothesis (\(H_0\)): The population mean is equal to 18.59 thousand miles, which means people are not driving less. - Alternative Hypothesis (\(H_a\)): The population mean is less than 18.59 thousand miles, indicating people are driving less.
02

Select the significance level

The problem has defined a significance level of \(\alpha = 0.05\), which will be used to judge the test's strength.
03

Calculate the test statistic

We need to calculate the t-statistic using the sample mean, population mean, sample standard deviation, and sample size.The formula for the t-statistic is:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Where:\(\bar{x} = 16.23\) thousand miles (sample mean),\(\mu = 18.59\) thousand miles (population mean),\(s = 4.06\) thousand miles (sample standard deviation),\(n = 15\) (sample size).Substitute these values into the formula:\[t = \frac{16.23 - 18.59}{4.06/\sqrt{15}} \approx -2.2875\]
04

Determine the critical value

For a one-tailed t-test with 14 degrees of freedom (\(n - 1 = 15 - 1 = 14\)) at a \(5\%\) significance level, we look up the critical value in the t-distribution table or use a calculator. The critical t-value is about \(-1.761\).
05

Compare the test statistic to the critical value

Compare the calculated t-statistic (\(-2.2875\)) to the critical value (\(-1.761\)). Since \(-2.2875 < -1.761\), the t-statistic is in the rejection region.
06

Make a decision

Because the t-statistic is less than the critical value, we reject the null hypothesis \(H_0\). This suggests that the average number of driving miles is indeed less than 18.59 thousand miles.

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Key Concepts

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

Understanding the t-test
The t-test is a statistical tool that helps us decide if there are significant differences between the means of two groups or between a sample mean and a known population mean. It is particularly useful when dealing with small sample sizes, like in this problem, where we have only 15 households.

To perform a t-test, we need some key components: the sample mean, the population mean, the sample standard deviation, and the sample size. These elements allow us to calculate the t-statistic, which is a kind of "distance" measure between the sample mean and the population mean.

It's important to check that the underlying data follows a normal distribution, which is an assumption when using the t-test. If this condition is met, the t-test can give us reliable results about the differences we are investigating.
Concept of the Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is the assumption that there is no effect or difference. It is the default position that there is no change from the status quo. In the context of our problem, the null hypothesis states that the average miles driven per household has not changed, remaining at 18.59 thousand miles.

The purpose of hypothesis testing is to challenge the null hypothesis. We seek evidence against it, to potentially reject it, rather than to prove it right. If our test shows significant evidence against \(H_0\), we may reject it and consider the alternative hypothesis.
Alternative Hypothesis Explained
The alternative hypothesis, symbolized as \(H_a\), represents what we are testing against the null hypothesis. It suggests that there is a change, effect, or difference. In this scenario, the alternative hypothesis claims that people are driving less than before, meaning the population mean is less than 18.59 thousand miles.

This hypothesis is typically what the researcher wants to prove with the data. If evidence from our test is strong enough to reject the null hypothesis, the alternative hypothesis is supported. An alternative hypothesis can be one-tailed or two-tailed, depending on the direction of the predicted effect. Here, a one-tailed test is used because we are specifically interested in whether people are driving less.
Significance Level Basics
The significance level, denoted as \(\alpha\), is a threshold we set to determine whether to reject the null hypothesis. It reflects the probability of rejecting a true null hypothesis, also known as making a Type I error. The significance level chosen for our problem is \(0.05\), or 5%.

By setting \(\alpha = 0.05\), we are allowing ourselves a 5% risk of concluding that people drive less, when in real terms they do not. The significance level helps us define the critical value or area in the statistical test, within which if our test statistic falls, we will reject the null hypothesis.

Choosing a significance level is a balance between risk and conservatism. A lower \(\alpha\) offers a more conservative test but increases the risk of not detecting a real effect, known as a Type II error.

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Most popular questions from this chapter

The mean increase in word family vocabulary among students in a one-year foreign language course is 576 word families. In order to estimate the effect of a new type of class scheduling, an instructor monitors the progress of 60 students; the sample mean increase in word family vocabulary of these students is 542 word families with sample standard deviation 18 word families. a. Test at the \(5 \%\) level of significance whether the mean increase with the new class scheduling is different from 576 word families, using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(5 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

An automobile manufacturer recommends oil change intervals of 3,000 miles. To compare actual intervals to the recommendation, the company randomly samples records of 50 oil changes at service facilities and obtains sample mean 3,752 miles with sample standard deviation 638 miles. Determine whether the data provide sufficient evidence, at the \(5 \%\) level of significance, that the population mean interval between oil changes exceeds 3,000 miles.

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \cdot \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average July temperature in a region historically has been \(74.5^{\circ} \mathrm{F}\). Perhaps it is higher now. b. The average weight of a female airline passenger with luggage was 145 pounds ten years ago. The FAA believes it to be higher now. c. The average stipend for doctoral students in a particular discipline at a state university is \(\$ 14,756\). The department chairman believes that the national average is higher. d. The average room rate in hotels in a certain region is \(\$ 82.53 .\) A travel agent believes that the average in a particular resort area is different. e. The average farm size in a predominately rural state was 69.4 acres. The secretary of agriculture of that state asserts that it is less today.

The recommended daily allowance of iron for females aged \(19-50\) is \(18 \mathrm{mg} /\) day. A careful measurement of the daily iron intake of 15 women yielded a mean daily intake of \(16.2 \mathrm{mg}\) with sample standard deviation 4.7 \(\mathrm{mg} .\) a. Assuming that daily iron intake in women is normally distributed, perform the test that the actual mean daily intake for all women is different from \(18 \mathrm{mg} /\) day, at the \(10 \%\) level of significance. b. The sample mean is less than 18, suggesting that the actual population mean is less than 18 \(\mathrm{mg} /\) day. Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Five years ago \(3.9 \%\) of children in a certain region lived with someone other than a parent. A sociologist wishes to test whether the current proportion is different. Perform the relevant test at the \(5 \%\) level of significance using the following data: in a random sample of 2,759 children, 119 lived with someone other than a parent.

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