
Concept explainers
A survey of 1000 households was taken to determine how they obtained news about current events. The survey considered only television.
Newspapers, and the Internet as sources for news. Of the households surveyed,
724 obtained news from television.
545 obtained news from newspapers.
280 obtained news from the Internet.
412 obtained news from both television and newspapers.
185 obtained news from both television and the Internet.
105 obtained news from television, newspapers, and the Internet.
64 obtained news from the Internet but not from television or newspapers. Of those households that were surveyed,
a. how many obtained news from television hut not from newspapers or the Internet?
b. how many obtained news from newspapers but not from television or the Internet?
c. how many obtained news from television or newspapers?
d. how many did not acquire news from television, newspapers, or the Internet?

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Chapter 2 Solutions
Mathematical Excursions (Looseleaf)
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- The average miles per gallon for a sample of 40 cars of model SX last year was 32.1, with a population standard deviation of 3.8. A sample of 40 cars from this year’s model SX has an average of 35.2 mpg, with a population standard deviation of 5.4. Find a 99 percent confidence interval for the difference in average mpg for this car brand (this year’s model minus last year’s).Find a 99 percent confidence interval for the difference in average mpg for last year’s model minus this year’s. What does the negative difference mean?arrow_forwardA special interest group reports a tiny margin of error (plus or minus 0.04 percent) for its online survey based on 50,000 responses. Is the margin of error legitimate? (Assume that the group’s math is correct.)arrow_forwardSuppose that 73 percent of a sample of 1,000 U.S. college students drive a used car as opposed to a new car or no car at all. Find an 80 percent confidence interval for the percentage of all U.S. college students who drive a used car.What sample size would cut this margin of error in half?arrow_forward
- You want to compare the average number of tines on the antlers of male deer in two nearby metro parks. A sample of 30 deer from the first park shows an average of 5 tines with a population standard deviation of 3. A sample of 35 deer from the second park shows an average of 6 tines with a population standard deviation of 3.2. Find a 95 percent confidence interval for the difference in average number of tines for all male deer in the two metro parks (second park minus first park).Do the parks’ deer populations differ in average size of deer antlers?arrow_forwardSuppose that you want to increase the confidence level of a particular confidence interval from 80 percent to 95 percent without changing the width of the confidence interval. Can you do it?arrow_forwardA random sample of 1,117 U.S. college students finds that 729 go home at least once each term. Find a 98 percent confidence interval for the proportion of all U.S. college students who go home at least once each term.arrow_forward
- Suppose that you make two confidence intervals with the same data set — one with a 95 percent confidence level and the other with a 99.7 percent confidence level. Which interval is wider?Is a wide confidence interval a good thing?arrow_forwardIs it true that a 95 percent confidence interval means you’re 95 percent confident that the sample statistic is in the interval?arrow_forwardTines can range from 2 to upwards of 50 or more on a male deer. You want to estimate the average number of tines on the antlers of male deer in a nearby metro park. A sample of 30 deer has an average of 5 tines, with a population standard deviation of 3. Find a 95 percent confidence interval for the average number of tines for all male deer in this metro park.Find a 98 percent confidence interval for the average number of tines for all male deer in this metro park.arrow_forward
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