Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
3rd Edition
ISBN: 9781259969454
Author: William Navidi Prof.; Barry Monk Professor
Publisher: McGraw-Hill Education
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Chapter 14, Problem 8CQ
To determine
To explain:Whether to reject the hypothesis of no interaction or not.
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Exercise 6.28 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insucient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents.
a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. Use Oregon as Group A and California as Group B.Find the test statistic and p-value.
b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?
In Exercises 17–20, refer to the sample of body temperatures (degrees Fahrenheit) in the table below. (The body temperatures are from a data set in Appendix B.)
Conclusion If we analyze the listed body temperatures with suitable methods of statistics, we conclude that when the differences are found between the 8 AM body temperatures and the 12 AM body temperatures, there is a 64% chance that the differences can be explained by random results obtained from populations that have the same 8 AM and 12 AM body temperatures. What should we conclude about the statistical significance of those differences?
In Exercises 1-5, refer to the following list of departure delay times (min) of American Airline flights from JFK airport in New York to LAX airport in Los Angeles. Assume that the data are samples randomly selected from larger populations.
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XLSTAT
Chapter 14 Solutions
Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
Ch. 14.1 - In Exercises 7 and 8, fill in each blank with the...Ch. 14.1 - Prob. 8ECh. 14.1 - Prob. 9ECh. 14.1 - Prob. 10ECh. 14.1 - Prob. 11ECh. 14.1 - Prob. 12ECh. 14.1 - In a one-way ANOVA, the following data were...Ch. 14.1 - In a one-way ANOVA, the following data were...Ch. 14.1 - Samples were drawn from three populations. The...Ch. 14.1 - Samples were drawn from three populations. The...
Ch. 14.1 - Pesticide danger: One of the factors that...Ch. 14.1 - Life-saving drug: Penicillin is produced by the...Ch. 14.1 - Pesticide danger: Using the data in Exercise 17,...Ch. 14.1 - Life-saving drug: Using the data in Exercise 18,...Ch. 14.1 - Artificial hips: Artificial hip joints consist of...Ch. 14.1 - Floods: Rapid drainage of floodwater is crucial to...Ch. 14.1 - Artificial hips: Using the data in Exercise 21,...Ch. 14.1 - Floods: Using the data in Exercise 22, perform the...Ch. 14.1 - Polluting power plants: Power plants can emit...Ch. 14.1 - Prob. 26ECh. 14.1 - Prob. 27ECh. 14.1 - Prob. 28ECh. 14.1 - Prob. 29ECh. 14.1 - Prob. 30ECh. 14.1 - Prob. 31ECh. 14.2 - Prob. 5ECh. 14.2 - Prob. 6ECh. 14.2 - Prob. 7ECh. 14.2 - Prob. 8ECh. 14.2 - In a two-way ANOVA, the P-value for interactions...Ch. 14.2 - Prob. 10ECh. 14.2 - Prob. 11ECh. 14.2 - Prob. 12ECh. 14.2 - Prob. 13ECh. 14.2 - Prob. 14ECh. 14.2 - Prob. 15ECh. 14.2 - Prob. 16ECh. 14.2 - Prob. 17ECh. 14.2 - Prob. 18ECh. 14.2 - Prob. 19ECh. 14.2 - Prob. 20ECh. 14.2 - Prob. 21ECh. 14.2 - Prob. 22ECh. 14.2 - Prob. 23ECh. 14.2 - Strong beams: The following table presents...Ch. 14.2 - Prob. 25ECh. 14.2 - Prob. 26ECh. 14.2 - Fruit yields: An agricultural scientist performed...Ch. 14.2 - Prob. 28ECh. 14.2 - Prob. 29ECh. 14 - Exercises 1-4 refer to the following data: At a...Ch. 14 - Prob. 2CQCh. 14 - Prob. 3CQCh. 14 - Prob. 4CQCh. 14 - Prob. 5CQCh. 14 - Prob. 6CQCh. 14 - Prob. 7CQCh. 14 - Prob. 8CQCh. 14 - Prob. 9CQCh. 14 - Prob. 10CQCh. 14 - Prob. 1RECh. 14 - Prob. 2RECh. 14 - Prob. 3RECh. 14 - Prob. 4RECh. 14 - Prob. 5RECh. 14 - Prob. 6RECh. 14 - Prob. 7RECh. 14 - Prob. 8RECh. 14 - Prob. 9RECh. 14 - Prob. 10RECh. 14 - Prob. 11RECh. 14 - Prob. 12RECh. 14 - Prob. 13RECh. 14 - Prob. 14RECh. 14 - Prob. 15RECh. 14 - Prob. 1WAICh. 14 - Prob. 2WAICh. 14 - Prob. 3WAICh. 14 - Prob. 4WAICh. 14 - Prob. 5WAICh. 14 - Prob. 1CSCh. 14 - Prob. 2CSCh. 14 - Prob. 3CSCh. 14 - Prob. 4CS
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- In a population-based cohort study, an entire community was interviewed regarding smoking habits and then followed for one year. Upon ascertainment of all lung cancer deaths, the investigator obtained the following data: Number of Individuals Lung Cancer Deaths Smokers 24,500 15 Nonsmokers 10,500 2 Calculate the risk difference per 100,000 per year. Round to the tenth decimaarrow_forwardA Bloomberg Businessweek North American subscriber study collected data from asample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annualincome of $75,000 or more, and 50% reported having an American Express credit card.a. What is the population of interest in this study?b. Is annual income a categorical or quantitative variable?c. Is ownership of an American Express card a categorical or quantitative variable?d. Does this study involve cross-sectional or time series data?e. Describe any statistical inferences Bloomberg Businessweek might make on the basisof the surveyarrow_forwardIn Exercises 1-5, refer to the following list of departure delay times (min) of American Airline flights from JFK airport in New York to LAX airport in Los Angeles. Assume that the data are samples randomly selected from larger populations. Normal Quantile Plot The accompanying normal quantile plot was obtained from the Flight 19 departure delay times. What does this graph tell us?arrow_forward
- 1. Observational studies suggest that moderate use of alcohol reduces heart attacks, and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks. In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two week period. Here are the percent changes in level for the subjects in both groups.Red 3.5 8.1 7.4 4.0 0.7 4.9 8.4 7.0 5.5White 3.1 0.5 -3.8 4.1 -0.6 2.7 1.9 -5.9 0.1 a. Is there good evidence that red wine drinkers gain more polyphenols on average than white wine drinkers? b. Does this study give reason to think that it is drinking red wine, rather than some lurking variable, that causes the increase in blood polyphenols? Justify your answer.c. Calculate and interpret a 95% confidence interval for…arrow_forwardA telephone company claims that the service calls which they receive are equally distributed among the five working days of the week. A survey of 110 randomly selected service calls was conducted. Is there enough evidence to refute the telephone company's claim that the number of service calls does not change from day-to-day? Days of the Week Mon Tue Wed Thu Fri Number of Calls 24 26 17 20 23 Copy Data Step 2 of 10: What does the null hypothesis indicate about the proportions of service calls received each day? Step 3 of 10: State the null and alternative hypothesis in terms of the expected proportions for each category. Step 4 of 10: Find the expected value for the number of service calls received on Monday. Round your answer to two decimal places. Step 5 of 10: Find the expected value for the number of service calls received on Tuesday. Round your answer to two decimal places. Step 6 of 10: Find the value of the test statistic. Round your answer to three decimal…arrow_forwardExercise 4.39 describes a study to investigate whether a recorded phone call is more effective than a flyer in persuading voters to vote for a particular candidate. The response variable is the proportion of voters planning to vote for the candidate, with pc and pf representing the proportions for the two methods (receiving a phone call and receiving a flyer, respectively). The sample statistic of interest is p^c−p^f. We are testing H0 : pc = pf vs Ha : pc > pf. A randomization distribution of differences in proportions, p^c−p^f, for this test is shown in Figure 4.18. Figure 4.18 Randomization distribution using n = 1000 for testing H0 : pc = pf a) Sketch a smooth curve that roughly approximates the distribution in Figure 4.18 and shade in the proportion of the area corresponding to the p-value for the sample statistic p^c−p^f=0.3. (b) Four possible sample statistics are given, along with four possible p-values. Match the statistics with the p-values.…arrow_forward
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