Question

3. During December 17-22, 2009, a CBS News poll surveyed a sample of adult Americans. The 95% confidence interval for the proportion of Americans who oppose putting a special tax on junk food such as soda, chips, and candy was (0.57, 0.63). a. What proportion of the adults in the sample opposed a special tax on junk food? [2] b. What is the margin of error of this confidence interval? (2) C. Which of the following could be a 99% confidence interval using the same data? Explain your choice. [2] i. (0.58, 0.62) ii. (0.56, 0.64) iii. (0.52, 0.57)

Answer 1

The $\small (1-\alpha)100\%$ confidence interval for the population proportion is given by $\small \frac{x}{n}-z_{\alpha/2}\sqrt{\frac{\frac{x}{n}(1-\frac{x}{n})}{n}}<p<\frac{x}{n}&plus;z_{\alpha/2}\sqrt{\frac{\frac{x}{n}(1-\frac{x}{n})}{n}}$ where $\frac{x}{n}$ is sample proportion

a)

Given 99% confidence interval as (0.57, 0.63)

Sample proportion = (0.57+0.63)/2 = 0.6

Therefore the proportion of adults in the sample who opposed a special tax on junk food is 60%

b)

Margin of error = 0.5*(width of confidence interval) = 0.5*(0.63-0.57) = 0.03

c)

95% confidence interval ==> $\small \alpha=0.05$

$\small z_{\alpha/2}$ for $\small \alpha=0.05$ is 1.96

99% confidence interval ==> $\small \alpha=0.01$

$\small z_{\alpha/2}$ for $\small \alpha=0.01$ is 2.575

Margin of error for 95% confidence interval

100 = = = = =

Margin of error for 99% confidence interval

= ANAL-4 = 2575 0.03 - 2575 - .04

Therefore 99% confidence interval is (0.6-0.04, 0.6+0.04) = (0.56, 0.64)

Correct option is (ii)