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Chapter 10: Problem 84

Seat belts help prevent injuries in automobile accidents, but they certainly don't offer complete protection in extreme situations. A random sample of 319 front-seat occupants involved in head-on collisions in a certain region resulted in 95 people who sustained no injuries ("Influencing Factors on the Injury Severity of Restrained Front Seat Occupants in Car-to-Car Head-on Collisions," Accident Analysis and Prevention \([1995]: 143-150\) ). Does this suggest that the true (population) proportion of uninjured occupants exceeds .25? State and test the relevant hypotheses using a significance level of \(.05\).

Short Answer

Expert verified
The exact conclusion will depend on the computed P-value and the decision made in Step 4, but in general, the conclusion would be that there is sufficient/not sufficient evidence at the 0.05 level of significance to conclude that the population proportion of uninjured occupants in head-on collisions exceeds 0.25.

Step by step solution

01

State the Hypotheses

The null hypothesis, denoted by \(H_0\), is that the population proportion \(p\) is equal to 0.25. The alternative hypothesis, denoted by \(H_A\), is that the population proportion \(p\) is greater than 0.25.\nSo, \(H_0: p = 0.25\)\n \(H_A: p > 0.25\)
02

Compute the Test Statistic

The test statistic for testing a hypothesis about a population proportion \(p\) when the null hypothesis is \(H0: p = p0\) is \[Z = \frac{\hat{p} - p0}{\sqrt{\frac{p0(1 - p0)}{n}}}\] where \(\hat{p}\) is the sample proportion and \(n\) is the sample size. Here, \(\hat{p} = \frac{95}{319}\) and \(n = 319\). So, plug in values and compute the test statistic.
03

Find the P-value

The P-value is the probability that a standard normal random variable is more than the observed value of the test statistic, under the null hypothesis. You can find this by looking up the value of the test statistic in a standard normal table or by using technology.
04

Make a Decision

If the P-value is less than the significance level of 0.05, reject the null hypothesis. Otherwise, do not reject the null hypothesis.
05

State the Conclusion

Based on the decision in Step 4, state the conclusion in context: If the null hypothesis is rejected, there is sufficient evidence at the 0.05 level of significance to conclude that the population proportion of uninjured occupants in head-on collisions exceeds 0.25. If the null hypothesis is not rejected, there is not sufficient evidence at the 0.05 level to conclude that the population proportion exceeds 0.25.

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Key Concepts

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

Population Proportion
The concept of population proportion is central when dealing with hypothesis testing. To put it simply, the population proportion is the fraction of a population that exhibits a certain characteristic. In this exercise, the characteristic in focus is the number of individuals uninjured in head-on collisions, and the proportion is represented by a symbol `p`. For example, in the exercise, we want to find out if the true proportion of uninjured front-seat occupants in head-on collisions exceeds 0.25.

Population proportions help us understand the larger picture based on a smaller sample. Here, a sample of 319 occupants is used to make inferences about the entire population of drivers. By saying a sample, we mean a smaller group drawn from the broader population to represent it. Using the sample proportion, denoted \(\hat{p}\), calculated as \(\frac{95}{319}\), we can conduct a hypothesis test to make broader inferences.
Significance Level
The significance level in hypothesis testing determines the threshold for rejecting the null hypothesis. It is symbolized by \(\alpha\) and chosen before conducting the test.
  • A common choice for significance level is 0.05, indicating a 5% risk of concluding that a difference exists when there is none.
  • The significance level informs us about how strong or weak evidence must be against the null hypothesis to warrant its rejection.
In the given problem, a significance level of 0.05 is used to decide whether there is enough evidence to reject the null hypothesis (that the population proportion of uninjured occupants is 0.25) and support the alternative hypothesis (that this proportion is greater than 0.25). The choice of 0.05 strikes a balance between type I error (false positive) and making practical conclusions from the data.
P-value
The P-value in statistical hypothesis testing is crucial as it helps determine the strength of the results.
  • It represents the probability of observing a test statistic as extreme, or more extreme, than the one computed from the dataset, assuming the null hypothesis is true.
  • The P-value helps in making a decision about whether to reject the null hypothesis.
During hypothesis testing, if the P-value is less than or equal to the significance level (0.05 in this case), it provides strong evidence against the null hypothesis, prompting its rejection. On the other hand, a higher P-value suggests insufficient evidence to reject the null hypothesis.
Finding the P-value involves computing the test statistic, often represented by a Z-score in tests of proportions, and comparing it to values in the standard normal distribution. Understanding the P-value helps in accurately drawing conclusions from the statistical tests.

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Most popular questions from this chapter

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January \(25 .\) 2006) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

According to the article "Workaholism in Organizations: Gender Differences" (Sex Roles [1999]: \(333-346\) ), the following data were reported on 1996 income for random samples of male and female MBA graduates from a certain Canadian business school: \begin{tabular}{lccc} & \(\boldsymbol{N}\) & \(\overline{\boldsymbol{x}}\) & \(\boldsymbol{s}\) \\ \hline Males & 258 & \(\$ 133,442\) & \(\$ 131,090\) \\ Females & 233 & \(\$ 105,156\) & \(\$ 98,525\) \\ \hline \end{tabular} Note: These salary figures are in Canadian dollars. a. Test the hypothesis that the mean salary of male MBA graduates from this school was in excess of \(\$ 100,000\) in \(1996 .\) b. Is there convincing evidence that the mean salary for all female MBA graduates is above \(\$ 100,000 ?\) Test using \(\alpha=.10\) c. If a significance level of \(.05\) or \(.01\) were used instead of \(.10\) in the test of Part (b), would you still reach the same conclusion? Explain.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a \(z\) statistic $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a^{2}}\) \(\mu>150\), where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. c. The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha ?\) (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the true value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure \(10.5\) ) of the sampling distribution of \(\bar{x}\), and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160 ?\) g. If \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.8\), what is the appropriate conclusion? What type of error might have been made in reaching this conclusion?

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(\mathrm{df}=9, t=0.73\) b. Upper-tailed test, \(\mathrm{df}=10, t=-0.5\) c. Lower-tailed test, \(n=20, t=-2.1\) d. Lower-tailed test, \(n=20, t=-5.1\) e. Two-tailed test, \(n=40, t=1.7\)

Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{x}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8 ?\) when \(\mu=9.5 ?\)
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