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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 8.3.7

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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Step 1: Understand the problem. The claim is that smokers have a mean cotinine level greater than 2.84 ng/mL. This is a one-tailed hypothesis test since the claim specifies 'greater than.' The test statistic provided is t = 56.319, and the sample size is n = 902.
Step 2: Identify the degrees of freedom (df). For a t-test, the degrees of freedom are calculated as df = n - 1, where n is the sample size. Substitute n = 902 into the formula to find df.
Step 3: Use the test statistic (t = 56.319) and degrees of freedom (df) to find the P-value. You can either use statistical software or a t-distribution table (Table A-3). If using the table, locate the range of P-values corresponding to the given t-value and df.
Step 4: Interpret the P-value. If the P-value is less than the significance level (commonly α = 0.05), reject the null hypothesis. Otherwise, fail to reject the null hypothesis. The null hypothesis states that the mean cotinine level for smokers is less than or equal to 2.84 ng/mL.
Step 5: Draw the final conclusion. Based on the P-value and the significance level, determine whether there is sufficient evidence to support the claim that smokers have a mean cotinine level greater than 2.84 ng/mL.

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Key Concepts

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

P-value

The P-value is a statistical measure that helps determine the significance of results from a hypothesis test. It represents the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true. A smaller P-value indicates stronger evidence against the null hypothesis, often leading researchers to reject it in favor of the alternative hypothesis.
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Step 3: Get P-Value

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) and an alternative hypothesis (H1), then using sample data to determine whether to reject H0. The process includes calculating a test statistic, finding the P-value, and comparing it to a predetermined significance level (alpha) to draw conclusions.
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Step 1: Write Hypotheses

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It measures how far the sample statistic is from the null hypothesis value, expressed in terms of standard errors. In this case, the test statistic is t = 56.319, which indicates a significant difference from the hypothesized mean cotinine level for nonsmokers, suggesting that smokers likely have a higher mean cotinine level.
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Step 2: Calculate Test Statistic
Related Practice
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?


Weights of Quarters The claim is that weights (grams) of quarters made after 1964 have a mean equal to 5.670 g as required by mint specifications. The sample size is and the test statistic is t = -3.135.

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Textbook Question

Using Technology

In Exercises 5–8, identify the indicated values or interpret the given display. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. Use a 0.05 significance level and answer the following:


a. Is the test two-tailed, left-tailed, or right-tailed?

b. What is the test statistic?

c. What is the P-value?

d. What is the null hypothesis, and what do you conclude about it?

e. What is the final conclusion?


Adverse Reactions to Drug The drug Lipitor (atorvastatin) is used to treat high cholesterol. In a clinical trial of Lipitor, 47 of 863 treated subjects experienced headaches (based on data from Pfizer). The accompanying TI-83/84 Plus calculator display shows results from a test of the claim that fewer than 10% of treated subjects experience headaches.

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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

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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Textbook Question

Randomization: Testing a Claim About a Proportion

In Exercises 5–8, use the randomization procedure for the indicated exercise.

Section 8-2, Exercise 9 “Cursed Movie”

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