Chapter 8.2, Problem 39E

a)

To determine

To identify the population and parameter.

Answer to Problem 39E

The population is all seniors in Tony’s school and parameter is proportion of seniors in Tony’s school that were planning to go to the prom.

Explanation of Solution

Given:

  $\begin{array}{l}Population=750\\ n=50\\ x=36\end{array}$

First need to understand about population and parameter.

Population: It is the set of all the possible individuals possessing the characteristic of interest in a study.

Parameter: A parameter is a numerical characteristic based on observations from the entire population of objects in a study.

In this study, population is all seniors in Tony’s school and parameter is proportion of seniors in Tony’s school that were planning to go to the prom.

b)

To determine

To explain whether the conditions for calculating a confidence interval for the population proportion.

Answer to Problem 39E

All the conditions are met.

Explanation of Solution

Given:

  $\begin{array}{l}Population=750\\ n=50\\ x=36\end{array}$

The condition of random selection is satisfied because the sample is SRS.

The sample should be less than 10% of the population. Here, 50 seniors are less than 10% of all the 750 seniors in population. Hence, this condition met.

The last condition is, number of success and failure must be $\ge 10$ . Therefore, number of successes =36 $\ge 10$ and number failures = 50-36=14 $\ge 10$ which is met. Hence, all the conditions of confidence interval are met.

c)

To determine

To construct 90% confidence interval for a proportion.

Answer to Problem 39E

The 90% confidence interval for a proportion is 0.6155 < p < 0.8245

Explanation of Solution

Given:

  $\begin{array}{l}Population=750\\ n=50\\ x=36\end{array}$

Confidence level = 0.90

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

The confidence interval:

  $(\widehat{p}-E,\widehat{p}+E)$

The confidence level = 0.90

So, level of significance = a=0.10

The z_c=z _a/2 critical value = 1.645 …Using excel formula, =ABS(NORMSINV(0.10/2))

The sample proportion is,

  $\widehat{p}=\frac{36}{50}=0.72$

The margin of error is,

  $\begin{array}{l}E=1.645\times \sqrt{\frac{0.72(1-0.72)}{50}}\\ E=0.1045\end{array}$

The confidence interval is,

  $\begin{array}{l}(0.72-0.1045,0.72+0.1045)\\ (0.6155,0.8245)\end{array}$

Hence, the 90% confidence interval for population proportion is 0.6155< p < 0.8245

d)

To determine

To interpret confidence interval.

Answer to Problem 39E

We are 90% confident that the true population proportion is between 0.6155 and 0.8245

Explanation of Solution

Given:

The 90% confidence interval for population proportion is 0.6155 < p < 0.8245

We are 90% confident that the true population proportion of seniors in Tony’s school that were planning to go to the prom is between0.6155 and 0.8245