Acollege 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

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=

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

Therefore,

n =

= 28.7875²

= 828.72 ≈ 829

Hence,

The correct answer is option (D) 829

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step-by-step explanation:

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a and c

step-by-step explanation: