Teens and their TV setsAccording to a Gallup Poll report, 64% of teens aged 13to 17have TVs in their rooms. Here is part of the footnote to this report:

These results are based on telephone interviews with a randomly selected national sample of 1028teenagers in the Gallup Poll Panel of households, aged 13to 17. For results based on this sample, one can say . . . that the maximum error attributable to sampling and other random effects is ±3percentage points. In addition to sampling error, question-wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls.16

(a) We omitted the confidence level from the footnote. Use what you have learned to determine the confidence level, assuming that Gallup took an SRS.

(b) Give an example of a “practical difficulty” that could lead to biased results for this survey.

Short Answer

Expert verified

From the given information,

a) The confidence level is 95%

b) Non response bias can lead to biased results for this survey.

Step by step solution

01

Part (a) Step 1: Given Information

It is given in the question that,

The margin of errorE=3%

Sample proportion p=64%

Sample size n=1028

We omitted the confidence level from the footnote. Use what you have learned to determine the confidence level, assuming that Gallup took an SRS.

02

Part (a) Step 2: Explanation

Sample proportion pis64%

Convert 64%into decimal.

64100=0.64

Now, calculate the margin of error. Use the formulaE=za/2×p^(1p^)n.

E=za/2×p^(1p^)n.

=zα/×0.64(10.64)1028(1)

=0.01497za2

Convert 3%into decimal.

3100=0.03

03

Part (a) Step 3: Explanation

Substitute 0.03for E in equation(1)

0.03=0.01497Za2

2a/2=0.030.01497

=2

Calculate the confidence level, Use table A.

P(2<Z<2)=P(Z<2)P(Z<2)

=0.9772-0.0228

=0.95

Therefore, the confidence level is 0.95.

Convert 0.95into percentage

0.95×100=95

Thus, the required confidence level is95.

04

Part (b) Step 1: Given Information

Give an example of a “practical difficulty” that could lead to biased results for this survey.

05

Part (b) Step 2: Explanation

There are three possible types of bias:

Selection bias will exclude part of the population.

Measurement of response bias will use a method that gives various values from the true value.

Nonresponse bias is the result of not having data for everybody in the sample

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Most popular questions from this chapter

Refer to Exercise 12. As Gallup indicates, the 3percentage point margin of error for this poll includes only sampling variability (what they call “sampling error”). What other potential sources of error (Gallup calls these “non-sampling errors”) could affect the accuracy of the59% estimate?

1. In the company’s prior-year survey, 80% of customers surveyed said they were “satisfied” or “very satisfied.” Using this value as a guess for pˆ, find the sample size needed for a margin of error of 3% at a 95% confidence level.

What if the company president demands 99% confidence instead? Determine how this would affect your answer to Question 1.

Teens' online profiles Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 487 teens with profiles found that 385 included photos of themselves. 13

(a) Construct and interpret a 95%confidence interval for p. Follow the four-step process.

(b) Is it plausible that the true proportion of American teens with profiles who have posted photos of themselves is 0.75? Use your result from part (a) to support your answer.

Explain briefly why each of the three conditions—Random, Normal, and Independent—is important when constructing a confidence interval.

A TV poll A television news program conducts a call-in poll about a proposed city ban on handgun ownership. Of the 2372 calls, 1921 oppose the ban. The station, following recommended practice, makes a confidence statement: "81%of the Channel 13 Pulse Poll sample opposed the ban. We can be 95% confident that the true proportion of citizens opposing a handgun ban is within1.6% of the sample result."

(a) Is the station's quoted 1.6% margin of error correct? Explain.

(b) Is the station's conclusion justifed? Explain.

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