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

829

Explanation:

Data provided in the question:

margin of error of E = $4

Confidence level = 90%

Standard deviation = $70

Now,

Sample size, n=
`\left ( (Z_(\alpha /2)* \sigma )/(E) \right )^(2)`

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

Therefore,

n =
`\left ( (1.645*70)/(4) \right )^(2)`

= 28.7875²

= 828.72 ≈ 829

Hence,

The correct answer is option (D) 829