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MA_775_DIABLO [31]
1 year ago
5

how many students must be randomly selected to estimate the mean weekly earnings of students at one college? we want 95% confide

nce that the sample mean is within $2 of the population mean, and the population standard deviation is known to be $10.
Mathematics
2 answers:
GREYUIT [131]1 year ago
7 0

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.

Learn more about estimation here:

brainly.com/question/24375372

#SPJ4

NISA [10]1 year ago
3 0

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 96.04

Step-by-step explanation:

The (1 - α)% confidence interval for population mean (μ) is:

X±Zₐ/2×\frac{σ}{\sqrt{n} }

The margin of error of a (1 - α)% confidence interval for population mean (μ) is:

MOE=Zₐ/2×\frac{σ}{\sqrt{n} }

The information provided is:

σ = $10

MOE = $2

The critical value of z for 95% confidence level is:

Zₐ/2=Z\left \atop {x=\frac{0.05}{2} }} \right.=1.96

Compute the sample size as follows:

MOE=Zₐ/2×\frac{σ}{\sqrt{n} }

n=[\frac{1.96*10}{2}]^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.

brainly.com/question/15685283

#SPJ4

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