Chapter 8.2, Problem 35E

(a)

To determine

To calculate: The population proportion when the confidence level is 99%.

Answer to Problem 35E

The population proportion is from 0.1833 to 0.2027 when the confidence interval is 99%.

Explanation of Solution

Given information:

Confidence level = 99%

Sample size = 2105

Total population = 10904

Formula used:

  $\text{Sample}\text{proportion}\text{=}\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}$

  $\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\text{α}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{×}\sqrt{\frac{\text{sample}\text{size}\left(\text{1}-\text{sample}\text{size}\right)}{\text{total}\text{population}}}$

Calculation:

Calculate the sample proportion use the formula.

  $\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{population}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{2105}{10904}\\ \approx 0.1930\end{array}$

Confidence level 99%.

Convert 99% into decimal.

  $\frac{99}{100}=0.99$

From the table,

  $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=z_{0.005}=2.575$

To calculate the margin of error,use the formula.

  $\begin{array}{c}\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\text{α}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{×}\sqrt{\frac{\text{sample}\text{size}\left(\text{1}-\text{sample}\text{size}\right)}{\text{total}\text{population}}}\\ E=z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sqrt{\frac{\stackrel{⌢}{p}\left(1-\stackrel{⌢}{p}\right)}{n}}\\ =2.575\times \sqrt{\frac{0.1930\left(1-0.1930\right)}{10904}}\\ \approx 0.0097\end{array}$

The sample proportion can vary by the amount margin of error. Thus, the confidence that contain true proportion also vary from minimum to maximum sample proportion.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.1930-0.0097<p<0.1930+0.0097\\ 0.1833<p<0.2027\end{array}$

Hence,the amount of confidence interval that contain the true population proportion is from 0.1833 to 0.2027.

(b)

To determine

Explain the article that a newspaper claims that 25% of U.S. college students are nondrinkers.

Explanation of Solution

Given information:

Confidence level = 99%

Sample size = 2105

Total population = 10904

From 35(a) population proportion is from 0.1833 to 0.2027 for the confidence level of 99%.

The newspaper article claims for 25% of sample population but confidence level does not contain the 25%.

Hence,there is a sufficient evidence to rejectthe newspaper claims that 25% of US college student are nondrinkers.