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Chapter 8: Problem 12

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Interstate Speeds It has been reported that the standard deviation of the speeds of drivers on Interstate 75 near Findlay, Ohio, is 8 miles per hour for all vehicles. A driver feels from experience that this is very low. A survey is conducted, and for 50 randomly selected drivers the standard deviation is 10.5 miles per hour. At \(\alpha=0.05,\) is the driver correct?

Short Answer

Expert verified
Yes, there is enough evidence to conclude the driver's belief is correct.

Step by step solution

01

Define Hypotheses

First, we define the null hypothesis and the alternative hypothesis. The null hypothesis (\( H_0 \)) assumes that the population standard deviation is the same as reported, 8 MPH. The alternative hypothesis (\( H_1 \)) suggests that the population standard deviation is greater than 8 MPH due to the driver's belief. Thus, we have:\[ H_0: \sigma = 8 \]\[ H_1: \sigma > 8 \]
02

Identify Test Statistic

Since the sample standard deviation is given and the sample size is 50, we will use the chi-square (\( \chi^2 \)) distribution for the test statistic calculation. The chi-square test statistic is given by:\[\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2}\]where \( n = 50 \) is the sample size, \( s = 10.5 \) is the sample standard deviation, and \( \sigma_0 = 8 \) is the population standard deviation under the null hypothesis.
03

Calculate Test Statistic

Substitute the values into the chi-square formula:\[\chi^2 = \frac{(50 - 1) \times (10.5)^2}{8^2} = \frac{49 \times 110.25}{64} \approx 84.3984\]
04

Determine Critical Value

For a right-tailed test with \( \alpha = 0.05 \) and 49 degrees of freedom (\( n - 1 = 49 \)), we can find the critical chi-square value from chi-square distribution tables. The critical value at \( \alpha = 0.05 \) and 49 degrees of freedom is approximately 66.338.
05

Make a Decision

Compare the calculated chi-square test statistic with the critical value. Since \( 84.3984 > 66.338 \), we reject the null hypothesis.
06

Conclusion

Since we rejected the null hypothesis, there is enough evidence at the \( \alpha = 0.05 \) significance level to conclude that the standard deviation of the speeds of drivers on Interstate 75 is greater than 8 MPH. Thus, the driver's belief is supported by the data.

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Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there's a significant difference between the expected and observed values in one or more categories. It's especially useful in analyzing categorical data, like frequencies in surveys or studies. By comparing the level of variance in data against what was expected, the Chi-Square Test helps to see if differences are due to random chance or indicate a pattern. In our exercise, we use the Chi-Square Test to assess whether the standard deviation of driver speeds is greater than what was reported.
Null Hypothesis
The Null Hypothesis, often denoted as \( H_0 \), is a foundational element in hypothesis testing. It is an assumption made for the sake of testing, stating there is no effect or difference in a particular situation. Essentially, it's a statement of equality or 'no change'.

For our exercise, the null hypothesis was that the standard deviation of vehicle speeds on Interstate 75 is 8 MPH. We tested this to determine if there was enough evidence to suggest any discrepancy from the reported standard deviation, serving as a benchmark throughout the testing process.
  • Denoted as \( H_0 \)
  • Assumes no change or effect
  • Serves as a starting point for the test
Alternative Hypothesis
The Alternative Hypothesis, labeled as \( H_1 \), stands in opposition to the null hypothesis. While the null suggests no change, the alternative is what the researcher aims to prove. It's often an indicator of a new effect or difference that counters the status quo.

In our case, the alternative hypothesis proposed that the standard deviation of driver speeds is greater than 8 MPH. It expresses the driver鈥檚 belief and is contrasted against the null hypothesis. Successfully rejecting the null hypothesis means acceptance of the alternative, suggesting there is indeed a larger standard deviation.
Significance Level
The Significance Level, usually expressed as \( \alpha \), is the criterion used to decide whether a result is statistically significant. It's the probability of rejecting the null hypothesis when it is actually true, acting as a threshold for decision-making. In hypothesis testing:

1. A common significance level is 0.05, meaning there is a 5% risk of incorrectly rejecting the null hypothesis.
2. If the p-value obtained from the test is less than \( \alpha \), the result is considered significant, leading to rejection of the null hypothesis.

For the exercise, we used \( \alpha = 0.05 \). This level helped determine whether the calculated chi-square statistic showed enough evidence to support the driver鈥檚 belief that the standard deviation was greater than the reported 8 MPH.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. High Temperatures in January Daily weather observations for southwestern Pennsylvania for the first three weeks of January for randomly selected years show daily high temperatures as follows: \(55,44,51,59,62,\) \(60,46,51,37,30,46,51,53,57,57,39,28,37,35,\) and 28 degrees Fahrenheit. The normal standard deviation in high temperatures for this time period is usually no more than 8 degrees. A meteorologist believes that with the unusual trend in temperatures the standard deviation is greater. At \(\alpha=0.05,\) can we conclude that the standard deviation is greater than 8 degrees?

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Exam Grades A statistics professor is used to having a variance in his class grades of no more than \(100 .\) He feels that his current group of students is different, and so he examines a random sample of midterm grades as shown. At \(\alpha=0.05,\) can it be concluded that the variance in grades exceeds \(100 ?\) $$ \begin{array}{lllll}{92.3} & {89.4} & {76.9} & {65.2} & {49.1} \\ {96.7} & {69.5} & {72.8} & {67.5} & {52.8} \\ {88.5} & {79.2} & {72.9} & {68.7} & {75.8}\end{array} $$

For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Takeout Food A magazine article reported that \(11 \%\) of adults buy takeout food every day. A fast-food restaurant owner surveyed 200 customers and found that 32 said that they purchased takeout food every day. At \(\alpha=0.02,\) is there evidence to believe the article's claim? Would the claim be rejected at \(\alpha=0.05 ?\)

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. $$ \begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array} $$

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Tornado Deaths A researcher claims that the standard deviation of the number of deaths annually from tornadoes in the United States is less than \(35 .\) If a random sample of 11 years had a standard deviation of 32, is the claim believable? Use \(\alpha=0.05 .\)
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