Question

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 99% confidence that the sample mean is within $4 of the population mean, and the population standard deviation is known to be $75

Answer 1

2333 students

Explanation:

Given:

Margin of error (E): 4

Standard deviation (σ): 75

Confidence level (CI): 99%

For this confidence level the value of z is equal to 2.576.

We can calculate the necessary sample size with the following formula:

`E=z\cdot\left((\sigma)/(√(n))\right)`

Substituting each value we calculate the value of n, like this:

`\begin{gathered} 4=2.576\cdot \left((75)/(√(n))\right) \\ \\ (75)/(√(n))=(4)/(2.576) \\ \\ √(n)=(75\cdot2.576)/(4) \\ \\ n=\left((75\cdot2.576)/(4)\right)^2 \\ \\ n=2332.89\cong2333 \end{gathered}`

The number of students to be selected must be 2333