Ch. 8 - Hypothesis Testing

Triola - Elementary Statistics 14th Edition

Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook

Triola 14th Edition

Ch. 8 - Hypothesis Testing

Problem 8.3.4

Chapter 8, Problem 8.3.4

Test Statistic and Critical Value The statistics for the sample data in Exercise 1 are n = 15, x_bar = 6.133333, and s = 8.862978, where the units are millions of dollars. Find the test statistic and critical value(s) for a test of the claim that the salaries are from a population with a mean greater than 5 million dollars. Assume that a 0.05 significance level is used.

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Step 1: Identify the hypothesis. The null hypothesis (H₀) is that the population mean μ is equal to 5 million dollars (H₀: μ = 5). The alternative hypothesis (H₁) is that the population mean μ is greater than 5 million dollars (H₁: μ > 5). This is a one-tailed test.

Step 2: Calculate the test statistic using the formula for the t-test: t = (x̄ - μ) / (s / √n). Here, x̄ is the sample mean (6.133333), μ is the hypothesized population mean (5), s is the sample standard deviation (8.862978), and n is the sample size (15). Substitute the given values into the formula.

Step 3: Determine the degrees of freedom (df) for the t-test. The degrees of freedom for a single-sample t-test is calculated as df = n - 1. For this problem, df = 15 - 1 = 14.

Step 4: Find the critical value for the t-test at a significance level of 0.05 for a one-tailed test. Use a t-distribution table or statistical software to find the critical t-value corresponding to df = 14 and α = 0.05.

Step 5: Compare the calculated test statistic to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis (H₀). Otherwise, fail to reject the null hypothesis.

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Key Concepts

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

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 mean is from the null hypothesis mean, expressed in terms of standard errors. In this context, the test statistic will help determine whether the sample mean salary significantly exceeds the hypothesized population mean of 5 million dollars.

Critical Value

The critical value is a threshold that determines the boundary for rejecting the null hypothesis in a statistical test. It is derived from the significance level (alpha), which in this case is 0.05. For a one-tailed test, the critical value indicates the point beyond which the test statistic would lead to rejecting the null hypothesis, thus supporting the claim that the mean salary is greater than 5 million dollars.

Significance Level

The significance level, denoted as alpha, is the probability of rejecting the null hypothesis when it is actually true, also known as a Type I error. In this scenario, a significance level of 0.05 implies that there is a 5% risk of concluding that the mean salary is greater than 5 million dollars when it is not. This level helps to control the likelihood of making erroneous conclusions based on the sample data.

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Step 4: State Conclusion Example 4

Related Practice

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.

Minting of Pennies Data Set 40 “Coin Weights” lists weights (grams) of pennies minted after 1983. Here are the statistics for those weights: n = 37, xbar = 2.49910 g, s = 0.01648 g . Use a 0.05 significance level to test the claim that the sample is from a population of pennies with weights having a standard deviation greater than 0.01000 g.

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Textbook Question

Estimates vs. Hypothesis Tests Labels on cans of Dr. Pepper soda indicate that they contain 12 oz of the drink. We could collect samples of those cans and accurately measure the actual contents, then we could use methods of Section 7-2 for making an estimate of the mean amount of Dr. Pepper in cans, or we could use those measured amounts to test the claim that the cans contain a mean of 12 oz. What is the difference between estimating the mean and testing a hypothesis about the mean?

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Textbook Question

Lightning Deaths Listed below are the numbers of deaths from lightning strikes in the United States each year for a sequence of recent and consecutive years. Find the values of the indicated statistics.

46 51 44 51 43 32 38 48 45 27 34 29 26 28 23 26 28 40 16 20

f. What important feature of the data is not revealed from an examination of the statistics, and what tool would be helpful in revealing it? What does a quick examination of the data reveal?

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Textbook Question

Final Conclusions

In Exercises 21–24, use a significance level of α = 0.05 and use the given information for the following:

State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.)

Without using technical terms or symbols, state a final conclusion that addresses the original claim.

Original claim: The mean pulse rate (in beats per minute) of adult males is 72 bpm. The hypothesis test results in a P-value of 0.0095.

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Textbook Question

Randomization: Testing a Claim About a Mean

In Exercises 9–12, use the randomization procedure for the indicated exercise.

Section 8-3, Exercise 23 “Cell Phone Radiation”

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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.

Internet Use A random sample of 5005 adults in the United States includes 751 who do not use the Internet (based on three Pew Research Center polls). Use a 0.05 significance level to test the claim that the percentage of U.S. adults who do not use the Internet is now less than 48%, which was the percentage in the year 2000. If there appears to be a difference, is it dramatic?

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