Respuesta :
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
How large of a sample of state employees should be taken if we want to estimate with 98% confidence the mean salary to be within $2,000? The population standard deviation is assumed to be $10,500. z-value for 98% confidence level is 2.326.
Answer:
Sample size = n = 150
Step-by-step explanation:
Recall that the margin of error is given by
[tex]$ MoE = z \cdot (\frac{\sigma}{\sqrt{n} } ) $\\\\[/tex]
Re-arranging for the sample size (n)
[tex]$ n = (\frac{z \cdot \sigma }{MoE})^{2} $[/tex]
Where z is the value of z-score corresponding to the 98% confidence level.
Since we want mean salary to be within $2,000, therefore, the margin of error is 2,000.
The z-score for a 98% confidence level is 2.326
So the required sample size is
[tex]n = (\frac{2.326 \cdot 10,500 }{2,000})^{2}\\\\n = (12.212)^{2}\\\\n = 149.13\\\\n = 150[/tex]
Therefore, we need to take a sample size of at least 150 state employees to estimate with 98% confidence the mean salary to be within $2,000.
