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Chapter 9: Problem 109

An April \(21,2009,\) USA Today article titled "On road, it's do as I say, not as I do" reported that \(58 \%\) of U.S. adults speed up to beat a yellow light. Suppose you conduct a survey in your hometown of 150 randomly selected adults and find that 71 out of the 150 admit to speeding up to beat a yellow light. Does your hometown have a lower rate for speeding up to beat a yellow light than the nation as a whole? Use a 0.05 level of significance.

Short Answer

Expert verified
Using these steps, the conclusion can be drawn whether your hometown has a lower rate of speeding up at a yellow light than the nation as a whole. The conclusion depends on the calculated P-value compared to the level of significance.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion for the hometown is equal to the proportion for the nation: \(H_0: p = 0.58\). The alternative hypothesis (\(H_a\)) is that the proportion for the hometown is lower than that for the nation: \(H_a: p < 0.58.\)
02

Compute the Test Statistic

The test statistic for a hypothesis test about a proportion is a Z-score. The formula for calculating the Z-score is: \(Z = (\widehat{p} - p_0) / \sqrt{ (p_0 \cdot (1-p_0)) / n }\), where \(\widehat{p}\) is the sample proportion, \(p_0\) is the national rate, and \(n\) is the sample size. So, \(Z = (71/150 - 0.58) / \sqrt{ (0.58 \cdot (1-0.58)) / 150 }.\)
03

Determine the P-Value

After computing the Z-score, use the standard normal table or Z-score calculator to determine the P-value. This step might require a statistical software or calculator.
04

Compare the P-value with the Level of Significance

If the calculated P-value is smaller than the level of significance, \(伪 = 0.05\), we reject the null hypothesis.
05

Make the Decision

If we rejected the null hypothesis, it means that the evidence supports the claim that the proportion in your hometown is lower than the national rate. If we failed to reject the null hypothesis, it means that there is not enough evidence to support that claim.

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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 a statement that there is no effect or no difference in a given experiment or statistical test. In the context of the exercise about beating yellow lights, the null hypothesis (H_0) asserts that the rate at which adults speed up to beat yellow lights in the hometown is the same as the national average (58%). It serves as the starting assumption for hypothesis testing and is denoted as H_0: p = 0.58. Testing this hypothesis involves gathering data, which in this case is the number of people who admit to this behavior, and analyzing it to determine whether there's enough evidence to reject the null hypothesis.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis (H_a) represents a statement that there is a difference or effect. For the yellow light scenario, the alternative hypothesis is that fewer adults in the survey participant's hometown speed up to beat yellow lights compared to the national rate: H_a: p < 0.58. One key point to note is that the alternative hypothesis is what the researcher is trying to prove, with the null hypothesis serving as the initial challenge to be disproven.
P-value
The p-value is a critical component of the hypothesis testing process, providing a measure of the strength of the evidence against the null hypothesis. The smaller the p-value, the stronger the evidence suggesting that the null hypothesis is not true. In this exercise, after calculating a test statistic, the p-value is determined by referring to the standard normal table or using a Z-score calculator. If this value is less than or equal to the significance level (0.05 in our exercise), it indicates that such extreme results are rare under the null hypothesis, leading to its rejection.
Significance Level
The significance level, denoted by 伪, is the threshold against which the p-value is compared to determine whether the null hypothesis should be rejected. It reflects the researcher's tolerance for Type I error, which is the incorrect rejection of a true null hypothesis. A common significance level used in many studies is 0.05. This means that there's a 5% chance of concluding that a difference exists when there is none鈥攁 risk that researchers are typically willing to accept to make a decision about the null hypothesis.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used as the basis for making a decision about whether to reject the null hypothesis. For proportion-based tests like the one in our exercise, the Z-score is commonly used as the test statistic. It indicates how many standard deviations the sample proportion is from the null hypothesis's claimed population proportion. In essence, it quantifies the discrepancy between the observed result and what we would expect if the null hypothesis were true. The calculation of the Z-score involves the sample proportion, the hypothesized population proportion, and the sample size.
Z-score
The Z-score, which is the test statistic in many hypothesis tests, is the number of standard deviations a data point is from the mean. When applied to hypothesis testing for proportions, as in the exercise provided, it involves subtracting the hypothesized population proportion from the sample proportion (the observed effect), dividing the difference by the standard error of the distribution. The computed Z-score reflects how extreme the sample result is鈥攊f it's a high or low value it could indicate that the sample provides sufficient evidence against the null hypothesis, depending on the direction of the alternative hypothesis.

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Most popular questions from this chapter

It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

Find \(n\) for a \(90 \%\) confidence interval for \(p\) with \(E=0.02\) using an estimate of \(p=0.25\).

A politician claims that she will receive \(60 \%\) of the vote in an upcoming election. The results of a properly designed random sample of 100 voters showed that 50 of those sampled will vote for her. Is it likely that her assertion is correct at the 0.05 level of significance? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

Even with a heightened awareness of beef quality, \(82 \%\) of Americans indicated their recent burger-eating behavior has remained the same, according to a recent T.G.I. Friday's restaurants random survey of 1027 Americans. In fact, half of Americans eat at least one beef burger each week. That's a minimum of 52 burgers each year. a. What is the point estimate for the proportion of all Americans who eat at least one beef burger per week? b. Find the \(98 \%\) confidence interval for the true proportion \(p\) in the binomial situation where \(n=1027\) and the observed proportion is one-half. c. Use the results of part b to estimate the percentage of all Americans who eat at least one beef burger per week.
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