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

Question

1 year ago

5

nce that the sample mean is within $2 of the population mean, and the population standard deviation is known to be $10.

2 answers:

GREYUIT [131] · Answer 1 · 2024-02-08T14:24:39+03:00

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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NISA [10] · Answer 2 · 2024-02-08T15:45:35+03:00

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.

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