Chapter 8.2, Problem 40E

a)

To determine

To identify the population and parameter.

Answer to Problem 40E

The population is all students who attend the school and parameter is proportion of students who the student body president knows the name of.

Explanation of Solution

Given:

  $\begin{array}{l}n=100\\ x=46\end{array}$

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 students who attend the school and parameter is proportion of students who the student body president knows the name of.

b)

To determine

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

Answer to Problem 40E

All the conditions are met.

Explanation of Solution

Given:

  $\begin{array}{l}n=100\\ x=46\end{array}$

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

The sample should be less than 10% of the population. Here, 100 students’ chips are less than 10% of all the 1800 students in population. Hence, this condition met.

The last condition is, number of success and failure must be $\ge 10$ . Therefore, number of successes =46 $\ge 10$ and number failures = 100-46=54 $\ge 10$ which is met. Hence, all the conditions of confidence interval are met.

c)

To determine

To construct 99% confidence interval for a proportion.

Answer to Problem 40E

The 99% confidence interval for a proportion is 0.3317 < p < 0.5883

Explanation of Solution

Given:

  $\begin{array}{l}n=100\\ x=46\end{array}$

Confidence level = 0.99

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 = 0.99

So, level of significance = a=0.01

The z_c=z _a/2 critical value = 2.575 …Using excel formula, =ABS(NORMSINV(0.01/2))

The sample proportion is,

  $\widehat{p}=\frac{46}{100}=0.46$

The margin of error is,

  $\begin{array}{l}E=2.575\times \sqrt{\frac{0.46(1-0.46)}{100}}\\ E=0.1283\end{array}$

The confidence interval is,

  $\begin{array}{l}(0.46-0.1283,0.46+0.1283)\\ (0.3317,0.5883)\end{array}$

Hence, the 99% confidence interval for population proportion is 0.3317 < p < 0.5883

d)

To determine

To interpret confidence interval.

Answer to Problem 40E

We are 99% confident that the true population proportion is between 0.3317 and 0.5883

Explanation of Solution

Given:

The 99% confidence interval for population proportion is 0.3317 < p < 0.5883

We are 99% confident that the true population proportion of all students that the students body president knows is 0.3317 and 0.5883