Chapter 8.2, Problem 33E

(a)

To determine

Explain the population and parameter of interest.

Answer to Problem 33E

All seniors of Tonya’s school are the total population for the sample and the population proportion of senior who plan to attend the prom is the parameter of interest.

Explanation of Solution

The population refers to the total set that are being studied. Thus all seniors of Tonya’s school is the total population for the sample.

The parameter of interest is the subject that was investigated. Here the parameter of interest is population proportion of senior who plan to attend the prom.

Hence, all seniors of Tonya’s school are the total population for the sample and the population proportion of senior who plan to attend the prom is the parameter of interest.

(b)

To determine

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

Answer to Problem 33E

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 seniors of the school.

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 36 and failure is 14. 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 90%.

Answer to Problem 33E

The population proportion is from 0.6155 to 0.8245 when the confidence interval is 90%.

Explanation of Solution

Given information:

Confidence level = 90%

Sample size $x=36$

Total population $n=50$

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

  $\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{36}{50}\\ =0.72\end{array}$

The confidence level is 90%.

Convert 90% into decimal.

  $\frac{90}{100}=0.90$

From the table,

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

To calculate 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}}\\ =1.645\times \sqrt{\frac{0.72\left(1-0.72\right)}{50}}\\ \approx 0.1045\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.72-0.1045<p<0.72+0.1045\\ 0.6155<p<0.8245\end{array}$

Hence,the amount of confidence interval that contain the true population proportion is from 0.6155 to 0.8245.

(d)

To determine

The confidence interval for the population proportion of seniors who attend the prom is 90%.

Answer to Problem 33E

The population proportion of seniors who attend the prom is from 0.6155 to 0.8245 when the confidence interval is 90%.

Explanation of Solution

Given information:

Confidence level = 90%

Sample size $x=36$

Total population $n=50$

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

  $\begin{array}{c}\text{Sample}\text{proportion}=\frac{\text{sample}\text{size}}{\text{total}\text{populatiom}}\\ \stackrel{⌢}{p}=\frac{x}{n}\\ =\frac{36}{50}\\ =0.72\end{array}$

The confidence level is 90%.

Convert 90% into decimal.

  $\frac{90}{100}=0.90$

From the table,

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

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}}\\ =1.645\times \sqrt{\frac{0.72\left(1-0.72\right)}{50}}\\ \approx 0.1045\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.

  $\begin{array}{l}\stackrel{⌢}{p}-E<p<\stackrel{⌢}{p}+E\\ 0.72-0.1045<p<0.72+0.1045\\ 0.6155<p<0.8245\end{array}$

Hence,the population proportion of seniors who attend the prom is from 0.6155 to 0.8245 when the confidence interval is 90%.