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Ch. 8 - Hypothesis Testing
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 8, Problem 8.4.19

Finding Critical Values of (chi)^2 For large numbers of degrees of freedom, we can approximate critical values of as follows:
(chi)^2 = (1/2)(z + sqrt(2k-1))


Here k is the number of degrees of freedom and z is the critical value(s) found from technology or Table A-2. In Exercise 12 “Spoken Words” we have df = 55, so Table A-4 does not list an exact critical value. If we want to approximate a critical value of (chi)^2 in the right-tailed hypothesis test with α = 0.01 and a sample size of 56, we let k =55 with z = 2.33 (or the more accurate value of z = 2.326348 found from technology). Use this approximation to estimate the critical value of for Exercise 12. How close is it to the critical value of (chi)^2 = 82.292 obtained by using Statdisk and Minitab?

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Step 1: Understand the formula for approximating the critical value of chi-squared: \( \chi^2 = \frac{1}{2}(z + \sqrt{2k - 1}) \). Here, \( k \) is the degrees of freedom, and \( z \) is the critical value obtained from technology or a table.
Step 2: Identify the given values from the problem. The degrees of freedom \( k \) is 55, and the critical value \( z \) is provided as 2.33 (or the more accurate value 2.326348).
Step 3: Substitute \( k = 55 \) into the formula \( \sqrt{2k - 1} \) to calculate \( \sqrt{2 \times 55 - 1} \). This simplifies to \( \sqrt{109} \).
Step 4: Substitute \( z = 2.33 \) (or \( z = 2.326348 \)) and \( \sqrt{109} \) into the formula \( \chi^2 = \frac{1}{2}(z + \sqrt{2k - 1}) \). This becomes \( \chi^2 = \frac{1}{2}(2.33 + \sqrt{109}) \) or \( \chi^2 = \frac{1}{2}(2.326348 + \sqrt{109}) \).
Step 5: Compare the approximated critical value of \( \chi^2 \) obtained using the formula to the exact critical value of \( \chi^2 = 82.292 \) provided in the problem. This step involves evaluating how close the approximation is to the exact value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chi-Squared Distribution

The Chi-Squared distribution is a statistical distribution commonly used in hypothesis testing, particularly in tests of independence and goodness-of-fit. It is defined by its degrees of freedom, which typically correspond to the number of categories or groups being analyzed. As the degrees of freedom increase, the distribution approaches a normal distribution, allowing for approximations in critical value calculations.
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Critical Value

A critical value is a threshold that determines the boundary for rejecting the null hypothesis in hypothesis testing. It is derived from the chosen significance level (α) and the relevant statistical distribution. In the context of the Chi-Squared distribution, the critical value indicates the point beyond which the observed statistic would lead to the rejection of the null hypothesis, thus providing a basis for decision-making.
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Degrees of Freedom

Degrees of freedom (df) refer to the number of independent values or quantities that can vary in a statistical calculation. In the context of the Chi-Squared test, degrees of freedom are typically calculated as the number of categories minus one. They play a crucial role in determining the shape of the Chi-Squared distribution and influence the critical values used in hypothesis testing.
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Related Practice
Textbook Question

Statistical Literacy and Critical Thinking

In Exercises 1–4, use the results from a Hankook Tire Gauge Index survey of a simple random sample of 1020 adults. Among the 1020 respondents, 86% rated themselves as above average drivers. We want to test the claim that more than 3/4 of adults rate themselves as above average drivers.


Requirements Are the requirements of the hypothesis test all satisfied? Explain.

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Textbook Question

Testing Claims About Proportions

In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.


Births A random sample of 860 births in New York State included 426 boys. Use a 0.05 significance level to test the claim that 51.2% of newborn babies are boys. Do the results support the belief that 51.2% of newborn babies are boys?

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Textbook Question

Testing Claims About Variation

In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.


Bank Lines The Jefferson Valley Bank once had a separate customer waiting line at each teller window, but it now has a single waiting line that feeds the teller windows as vacancies occur. The standard deviation of customer waiting times with the old multiple-line configuration was 1.8 min. Listed below is a simple random sample of waiting times (minutes) with the single waiting line. Use a 0.05 significance level to test the claim that with a single waiting line, the waiting times have a standard deviation less than 1.8 min. What improvement occurred when banks changed from multiple waiting lines to a single waiting line?


6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7

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Textbook Question

Testing Claims About Variation

In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Pulse Rates of Men A simple random sample of 153 men results in a standard deviation of 11.3 beats per minute (based on Data Set 1 “Body Data” in Appendix B). The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute; see the accompanying StatCrunch display for this test. What do the results indicate about the effectiveness of using the range rule of thumb with the “normal range” from 60 to 100 beats per minute for estimating in this case?

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Textbook Question

Testing Claims About Proportions

In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.


Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Should this be comforting to physicians?

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Textbook Question

Finding P-values

In Exercises 5–8, either use technology to find the P-value or use Table A-3 to find a range of values for the P-value. Based on the result, what is the final conclusion?


Cotinine in Smokers The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n = 902 and the test statistic is t = 56.319.

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