A population has a standard deviation $\sigma = 18.4$ How large a sample must be drawn so that a 98% confidence interval for μ will have a margin of error equal to 4.1?
A) 192
B) 5
C) 109
D) 10

Question

Quizplus · Accepted Answer

The margin of error is given by: 
$$ME = z_{\alpha/2} 	imes \dfrac{\sigma}{\sqrt{n}}$$

where,
$$z_{\alpha/2} = z_{0.01/2} = z_{0.005} \approx 2.33$$ (from standard normal distribution table)
Putting the given values in the above equation, we get:
$$4.1 = 2.33 	imes \dfrac{18.4}{\sqrt{n}}$$
Solving for n, we get:
$$n = \left(\dfrac{2.33 	imes 18.4}{4.1}ight)^2 \approx 109$$

Therefore, the sample size must be 109.

A Population Has a Standard Deviation σ=18.4\sigma = 18.4σ=18.4 How Large a Sample Must Be Drawn So That a Large

A Population Has a Standard Deviation $\sigma = 18.4$ How Large a Sample Must Be Drawn So That a Large