Question

A college would like to estimate the average monthly income of its students. How many students must be sampled in order to estimate mu with 90% confidence and a margin of error of E = $4? Suppose the population standard deviation is known to be
A. $70.
B. 270
C. 288
D. 829
E. 249

Answer 1

Answer:

829

Step-by-step explanation:

Data provided in the question:

margin of error of E = $4

Confidence level = 90%

Standard deviation = $70

Now,

Sample size, n= $\left ( \frac{Z_{\alpha /2}\times \sigma }{E} \right )^{2}$

for 90% confidence level, z value = 1.645            [from standard z-table]

Therefore,

n = $\left ( \frac{1.645\times70}{4} \right )^{2}$

= 28.7875²

= 828.72 ≈ 829

Hence,

The correct answer is option (D) 829