Question 1165659: The monthly earnings of a group of recent business school graduates are normally distributed with a standard deviation of 589 dollars. A researcher wants to estimate the mean monthly earnings of all recent graduates from the school. Find the sample size needed to have a confidence level of 95% and a margin of error of 132 dollars.
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! The sample size needed to estimate the mean monthly earnings is **77**.
To find the required sample size ($n$), we use the formula for estimating a population mean:
$$n = \left(\frac{Z \cdot \sigma}{ME}\right)^2$$
### 1. Identify the Variables
* **Standard Deviation ($\sigma$):** $\$589$
* **Margin of Error ($ME$):** $\$132$
* **Confidence Level:** $95\%$
* **Z-score ($Z$):** For a $95\%$ confidence level, the critical $Z$-score is $\mathbf{1.96}$.
### 2. Calculate the Sample Size
Substitute the values into the formula:
$$n = \left(\frac{1.96 \times 589}{132}\right)^2$$
First, calculate the term inside the parenthesis:
$$n = \left(\frac{1154.44}{132}\right)^2$$
$$n \approx (8.745757...)^2$$
$$n \approx 76.487$$
### 3. Round Up
Since the sample size must be a whole number and you must err on the side of caution (to ensure the margin of error is *no more* than \$132), you must **always round up** to the next whole number.
$$n = \mathbf{77}$$
The researcher needs a sample size of **77** graduates.
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