Chapter 8.2, Problem 36E

(a)

To determine

To calculate: The population proportion when the confidence level is 95%.

Answer to Problem 36E

The population proportion is 0.7545 to 0.8267 when the confidence interval is 90%.

Explanation of Solution

Given information:

Confidence level = 95%

Sample size = 385

Total population = 487

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

  $\begin{array}{c}\text{Sample}\text{proportion=}\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{385}{487}\\ \approx 0.7906\end{array}$

Confidence level 95%.

Convert 95% into decimal.

  $\frac{95}{100}=0.95$

From the table,

  $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=z_{0.025}=1.96$

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}}\\ =1.96\times \sqrt{\frac{0.7906\left(1-0.7906\right)}{487}}\\ \approx 0.0361\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.7906-0.0361<p<0.7906+0.0361\\ 0.7545<p<0.8267\end{array}$

Hence,the confidence interval that contain the true population proportionis from 0.7545 to 0.8267.

(b)

To determine

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

Answer to Problem 36E

The survey that claim the true proportion of American teens with profiles who have posted photos of themselves is 0.75 is rejected.

Explanation of Solution

Given information:

Confidence level = 95%

Sample size = 385

Total population = 487

From 36(a) population proportion is from 0.7574 to 0.8267 for the confidence level of 0.95.

Since the survey claims for 0.75 of sample population but the confidence level is 0.95.

Hence,the survey that claimthe true proportion of American teens with profiles who have posted photos of themselves is 0.75 is rejected.