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Chapter 7: Q. 2.6 (page 438)


On one episode of his show, a radio show host encouraged his listeners to visit his website and vote in a poll about proposed tax increases. Of the 4821 people who vote, 4277 are against the proposed increases. To which of the following populations should the results of this poll be generalized?

a. All people who have ever listened to this show

b. All people who listened to this episode of the show

c. All people who visited the show host's website

d. All people who voted in the poll

e. All people who voted against the proposed increases

Short Answer

Expert verified

d. All people who voted in the poll the results of this poll be generalized .

Step by step solution

01

Given Information

A radio show host encouraged his listeners to visit his website and vote in a poll over potential tax rises on one of his shows. 4277 persons out of 4821 voted against the proposed hikes.

02

Explanation for correct option

The general public is interested in a variety of topics about which they desire to learn more. A sample is a small portion of a larger population.

The sample is observed to be the 4821 persons that voted on the website. People with strong opinions are more likely to participate in a voluntary response sample, hence there will only be data about strong opinions, resulting in consistent overestimation and underestimation of genuine opinions.

As a result, the best solution is (d)

03

Explanation for incorrect option

a. All people who have ever listened to this show is not the answer.

b. All people who listened to this episode of the show is not the answer.

c. All people who visited the show host's website is not the answer.

e. All people who voted against the proposed increases is not the answer.

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Most popular questions from this chapter

The number of unbroken charcoal briquets in a 20-pound bag filled at the factory follows a Normal distribution with a mean of 450 briquets and a standard deviation of 20 briquets. The company expects that a certain number of the bags will be underfilled, so the company will replace for free the 5%of bags that have too few briquets. What is

the minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free?

a. 404

b. 411

c. 418

d. 425

e. 448

The mean of this distribution (don鈥檛 try to find it) will be

a. very close to the median.

b. greater than the median.

c. less than the median.

d. You can鈥檛 say, because the distribution isn鈥檛 symmetric.

e. You can鈥檛 say, because the distribution isn鈥檛 Normal.

An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is and the standard deviation of the loss is 蟽=\(5000.The distribution of losses is strongly right-skewed: many policies have \)0loss, but a few have large losses. The company hopes to sell 1000 of these policies for \(300each.

a. Assuming that the company鈥檚 claim is true, what is the probability that the mean loss from fire is greater than \)300for an SRS of 1000 homeowners?

b. If the company wants to be 90% certain that the mean loss from fire in an SRS of 1000 homeowners is less than the amount it charges for the policy, how much should the company charge?

Increasing the sample size of an opinion poll will reduce the

a. bias of the estimates made from the data collected in the poll.

b. variability of the estimates made from the data collected in the poll.

c. effect of nonresponse on the poll.

d. variability of opinions in the sample.

e. variability of opinions in the population.

A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has mean 1.6 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hour.

a. Without doing any calculations, explain which event is more likely:

  • randomly selecting 1 car entering this interchange during rush hour and finding 2 or more people in the car
  • randomly selecting 35 cars entering this interchange during rush hour and finding an average of 2 or more people in the cars

b. Explain why you cannot use a Normal distribution to calculate the probability of the first event in part (a).

c. Calculate the probability of the second event in part (a).
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