Exam 21: What Is a Confidence Interval
Exam 1: Where Do Data Come From30 Questions
Exam 2: Samples, Good and Bad30 Questions
Exam 3: What Do Samples Tell Us55 Questions
Exam 4: Sample Surveys in the Real World36 Questions
Exam 5: Experiments, Good and Bad50 Questions
Exam 6: Experiments in the Real World32 Questions
Exam 7: Data Ethics21 Questions
Exam 8: Measuring33 Questions
Exam 9: Do the Numbers Make Sense25 Questions
Exam 10: Graphs, Good and Bad30 Questions
Exam 11: Displaying Distributions With Graphs22 Questions
Exam 13: Normal Distributions54 Questions
Exam 14: Describing Relationships: Scatterplots and Correlation56 Questions
Exam 15: Describing Relationships: Regression, Prediction, and Causation37 Questions
Exam 16: The Consumer Price Index and Government Statistics31 Questions
Exam 17: Thinking About Chance25 Questions
Exam 18: Probability Models30 Questions
Exam 19: Simulation20 Questions
Exam 20: The House Edge: Expected Values30 Questions
Exam 21: What Is a Confidence Interval43 Questions
Exam 22: What Is a Test of Significance30 Questions
Exam 23: Use and Abuse of Statistical Inference18 Questions
Exam 24: Two-Way Tables and the Chi-Square Test47 Questions
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The student newspaper at a college asks a simple random sample of 250 undergraduates, "Do you favor eliminating supplemental fees for lab courses?" In all, 150 of the 250 are in favor of eliminating such fees.
To estimate p, you will use the proportion = 150/250 of your sample who favored eliminating supplemental fees for lab courses. The number is a
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The student newspaper at a college asks a simple random sample of 250 undergraduates, "Do you favor eliminating supplemental fees for lab courses?" In all, 150 of the 250 are in favor of eliminating such fees.
Suppose that (unknown to you) 55% of all undergraduates favor eliminating supplemental fees for lab courses. If you took a very large number of simple random samples of size n = 250 from this population, the sampling distribution of the sample proportion p would be approximately normal with
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In a simple random sample of 144 households in a city in Kentucky, the average number of children in these households was 1.12 children. The standard deviation from this sample was 2.40 children. A 90 percent confidence interval for the mean number of children in all households in this city is:
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