The sample of students required to estimate the mean weekly earnings of students at one college is of size 96.04.
For the population mean (μ) , we have the (1 - α)% confidence interval as:
X ± Zₐ / 2 + I / √n
margin of error = MOE = Zₐ / 2 ×I / √n
We are given:
σ = $10
MOE = $2
The critical value of z for 95% confidence level is
Zₐ / 2 = Zₓ = 1.96 ( for x as 0.025)
n = (1.96 (10))²
n = 96.04
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 96.04.
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Answer:
Step-by-step explanation:
3*3+2/3-2*6+5/6
=9+2/3-12+5/6
=11/3-17/6
take lcm of the denominator
=11*2-17*1/6
=22-17/6
=5/6
Answer:
The initial value is the y value when x is zero. Look at the table to help you determine the y value when x is zero and that is your answer.
Step-by-step explanation:
Do you still need an answer?
Answer:
8/9 sorry if wrong
Step-by-step explanation:
