91影视

It is given in the question that,

The margin of error$E=3\%$

Sample proportion $\stackrel{鈭�}{p}=64\%$

Sample size $n=1028$

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

Sample proportion $\stackrel{鈭�}{p}is64\%$

Convert $64\%$into decimal.

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

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

$E=za/2脳\sqrt{\frac{\hat{p}(1鈭�\hat{p})}{n}}.$

$=z伪/脳\sqrt{\frac{0.64(1鈭�0.64)}{1028}}鈰�鈰�鈥�\left(1\right)$

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

Convert $3\%$into decimal.

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

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.

$P(鈭�2<Z<2)=P(Z<2)鈭�P(Z<鈭�2)$

$=0.9772-0.0228$

$=0.95$

Therefore, the confidence level is $0.95$.

Convert $0.95$into percentage

$0.95脳100=95$

Thus, the required confidence level is$95$.

Give an example of a 鈥減ractical dif铿乧ulty鈥� that could lead to biased results for this survey.

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