Question

Teens and their TV setsAccording to a Gallup Poll report, $64$% of teens aged $13$to $17$have 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 $1028$teenagers in the Gallup Poll Panel of households, aged $13$to $17$. For results based on this sample, one can say . . . that the maximum error attributable to sampling and other random effects is $\pm 3$percentage 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

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

Part (a) Step 1: Given Information

It is given in the question that,

The margin of error$E=3\%$

Sample proportion $\stackrel{\wedge }{p}=64\%$

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.

Part (a) Step 2: Explanation

Sample proportion $\stackrel{\wedge }{p}is64\%$

Convert $64\%$into decimal.

$\frac{64}{100}=0.64$

Now, calculate the margin of error. Use the formula$E=za/2\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$

$E=za/2\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$

$=z\alpha /\times \sqrt{\frac{0.64(1-0.64)}{1028}}\cdots \cdots \dots \left(1\right)$

$=0.01497z\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Convert $3\%$into decimal.

$\frac{3}{100}=0.03$

Part (a) Step 3: Explanation

Substitute $0.03$for E in equation(1)

$0.03=0.01497Z\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$2a/2=\frac{0.03}{0.01497}$

$=2$

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 $0.95$into percentage

$0.95\times 100=95$

Thus, the required confidence level is$95$.

Part (b) Step 1: Given Information

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

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