Chapter 8.2, Problem 40E

To determine

To calculate: The population proportion when the confidence level is 90%.

Answer to Problem 40E

The population proportion is from 0.2276 to 0.2720 when the confidence interval is 90%.

Explanation of Solution

Given Information:

Number of success = 256

Number of failure = 769

Confidence interval = 90%

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

  $\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\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:

Step1 to construct the population proportion for 90% of confidence interval.

Step 2 check for all three conditions.

The condition for random is satisfied because all the students comes randomly.

Since the sample of 1025 students are less than 10% of the entire population of student.Thus, the independence condition is satisfied.

The number of success and number of failure are greater than 10. Thus, condition of normal is also satisfied.

Hence, all conditions are satisfied.

Step3 find population proportion

  $\begin{array}{c}\text{Sample}\text{size}=\text{number}\text{of}\text{success}+\text{number}\text{of}\text{failure}\\ =256+769\\ =1025\end{array}$

Calculate sample proportion use the formula$\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{256}{1025}\\ \approx 0.2498\end{array}$

The confidence interval is 90%.

Convert 90% into decimal.

  $\frac{90}{100}=0.90$

From table, $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=z_{0.05}=1.645$

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}}\\ =1.645\times \sqrt{\frac{0.2498\left(1-0.2498\right)}{1025}}\\ \approx 0.0222\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.2498-0.0222<p<0.2498+0.0222\\ 0.2276<p<0.2720\end{array}$

Step 4 90% of the confidence interval that contain the true population proportion is from 0.2276 to 0.2720.

Hence,the confidence interval that contain the true population proportion is from 0.2276 to 0.2720.