Chapter 9.3, Problem 9.42E

a)

To determine

To show that the 95% confidence interval for p is correct.

Answer to Problem 9.42E

The 95% confidence interval is correct.

Explanation of Solution

Given:

  $x$ = 1921

n = 2372

Confidence level = 0.95

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Calculation:

Sample proportion,

  $\widehat{p}=\frac{1927}{2372}=0.8124=81\%$

Now need to find Zc

The level of significance is = a =0.05

Therefore, Zc = Z _a/2 = 1.96 ..Using standard normal table

The margin of error is,

  $E=1.96\times \sqrt{\frac{0.8124(1-0.8124)}{2372}}=0.0157=1.6\%$

Therefore, sample proportion is 81% and margin of error is 1.6%, hence 95% confidence interval is correct.

b)

To determine

To explain whether the stations conclusion is not justified.

Answer to Problem 9.42E

The station's conclusion is not justified.

Explanation of Solution

(b) The sample is not a random sample as the sample is a call-in poll. Therefore, the station's conclusion is not justified. This means that people will have to voluntary call in their response, making the sample a voluntary response sample. People with the strongest believes tend to participate in a voluntary response sample and thus the sample will not be representative for the population.

However, the sample will result from nonresponse bias, because not everybody who hears/reads about the poll will actually call in their opinion.