Chapter 8.2, Problem 34E

(a)

To determine

Explain the population and parameter of interest.

Answer to Problem 34E

All undergraduate participants of the quiz are the total population for the sampleand the population proportion of all undergraduate participants of the quiz.

Explanation of Solution

The population refers to the total set that are being studied. Thus all undergraduate participants of the quiz are the total population for the sample.

The parameter of interest is the subject that was investigated. Here the parameter of interest is the population proportionof all undergraduate participants of the quiz.

Hence,all undergraduate participants of the quiz are the total population for the sampleand the population proportion of all undergraduate participants of the quiz.

(b)

To determine

Check whether the conditions for construction of a confidence interval for the parameter are satisfied or not.

Answer to Problem 34E

All three conditions for confidence interval are satisfied.

Explanation of Solution

To find a confidence interval three conditions should be satisfied. Conditions are Random, Independent and Normal.

The condition of random is satisfied since the sample was taken at randomly from the students of the university.

Since the sample size is less than 10% of the population size thus the condition for independent is satisfied.

In the survey the number of success is 19 and failure is 153. Since both the success and failure are greater than 10. Thus, condition for normal is satisfied.

Hence, all three conditions for confidence interval are satisfied.

(c)

To determine

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

Answer to Problem 34E

The population proportion is from 0.0489 to 0.1721 when the confidence interval is 99%.

Explanation of Solution

Given information:

Confidence level = 99%

Sample size $x=19$

Total population $n=172$

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:

Find the sample proportion use the formula $\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}$ .

  $\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{19}{172}\\ =0.1105\end{array}$

The confidence level is 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 find the margin of error, use the formula $\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\text{α}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{×}\sqrt{\frac{\text{sample}\text{size}\left(\text{1}-\text{sample}\text{size}\right)}{\text{total}\text{population}}}$ .

  $\begin{array}{c}\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\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{0.1105\left(1-0.1105\right)}{172}}\\ \approx 0.0616\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.1105-0.0616<p<0.1105+0.0616\\ 0.0489<p<0.1721\end{array}$

Hence,the amount of confidence interval that contain the true population proportion is from 0.0489 to 0.1721.

(d)

To determine

The confidence interval for the population proportion of students willing to report cheating by other students is 99%.

Answer to Problem 34E

The population proportion of seniors who attend the prom is from 0.0489 to 0.1721 when the confidence interval is 99%.

Explanation of Solution

Given information:

Confidence level = 99%

Sample size $x=19$

Total population $n=172$

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:

Find the sample proportion use the formula $\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}$ .

  $\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{19}{172}\\ =0.1105\end{array}$

The confidence level is 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 find the margin of error, use the formula $\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\text{α}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{×}\sqrt{\frac{\text{sample}\text{size}\left(\text{1}-\text{sample}\text{size}\right)}{\text{total}\text{population}}}$ .

  $\begin{array}{c}\text{Margin}\text{of}\text{error}={\text{z}}_{\raisebox{1ex}{$\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{0.1105\left(1-0.1105\right)}{172}}\\ \approx 0.0616\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.1105-0.0616<p<0.1105+0.0616\\ 0.0489<p<0.1721\end{array}$

Hence,the population proportion of students willing to report cheating by other students is from 0.0489 to 0.1721 when the confidence interval is 99%.