Question

The student body president of a high school claims to know the names of at least $1000$of the 1800 students who attend the school. To test this claim, the student government advisor randomly selects $100$students and asks the president to identify each by name. The president successfully names only $46$of the students.

a. Identify the population and parameter of interest.

b. Check conditions for constructing a confidence interval for the parameter.

c. Construct a $99\%$confidence interval for p.

d. Interpret the interval in context.

Short answer

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

Part b. All the conditions are met.

Part c. The $99\%$confidence interval for a proportion is $0.3317<p<0.5883$

Part d. We are $99\%$confident that the true population proportion is between$0.3317and0.5883$.

Step-by-step solution

Part a. Step 1. Given information

$$\begin{gathered} n=1000 \\ x=46 \end{gathered}$$

Part a. Step 2. Explanation

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.

Part b. Step 1. Explanation

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.

Part c. Step 1. Formula Used

Sample proportion:

$\hat{p}=\frac{x}{n}$

Margin of error:

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

The confidence interval:

$(\hat{p}-E,\hat{p}+E)$

Part c. Step 2. Explanation

The confidence level$=0.99$

So, level of significance$=a=0.01$

The $z_c=z_{\alpha /2}$critical value $=2.575$....

The sample proportion is,

$\hat{p}=\frac{46}{100}=0.46$

The margin of error is,

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

The confidence interval is,

role="math" localid="1663922962651" $$\begin{gathered} (0.46-0.1283,0.46+0.1283) \\ (0.3317,0.5883) \end{gathered}$$

Hence, the $99\%$confidence interval for population proportion isrole="math" localid="1663922956187" $0.3317<p<0.5883$

Part c. Step 1. Explanation

We are $99\%$confident that the true population proportion of all students that the students body president knows is$0.3317and0.5883$