Question

In: Math

A random sample of 40 students has a mean annual earnings of $3120


A random sample of 40 students has a mean annual earnings of $3120. 

Assume the population standard deviation, σ, is $677. (Section 6.1) 


• Construct a 95% confidence interval for the population mean annual earnings of students. 

 Margin of error, E._______  Confidence interval: _______ <μ< _______ 


• If the number of students sampled was reduced to 25 and the level of confidence remained at 95%, what would be the new error margin and confidence interval?

 Margin of error, E._______  Confidence interval: _______ <μ< _______ 


• Did the confidence interval increase or decrease and why? 


• If the number of students sampled was increased to 100 and the level of confidence remained at 95%, what would be the new confidence interval? 

 Margin of error, E._______  Confidence interval: _______ <μ< _______ 


• Did the confidence interval increase or decrease and why?

Solutions

Expert Solution

2)

a_

sample mean, xbar = 3120
sample standard deviation, σ = 677
sample size, n = 40


Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96


ME = zc * σ/sqrt(n)
ME = 1.96 * 677/sqrt(40)
ME = 209.8

margin of error = 209.8

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3120 - 1.96 * 677/sqrt(40) , 3120 + 1.96 * 677/sqrt(40))
CI = (2910.2 , 3329.8)

2910.20 < mu < 3329.80

b)

sample mean, xbar = 3120
sample standard deviation, σ = 677
sample size, n = 25


Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96


ME = zc * σ/sqrt(n)
ME = 1.96 * 677/sqrt(25)
ME = 265.38

margin of error = 265.38

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3120 - 1.96 * 677/sqrt(25) , 3120 + 1.96 * 677/sqrt(25))
CI = (2854.62 , 3385.38)

2854.62

c)

Increased because sample size is reduced

d)

sample mean, xbar = 3120
sample standard deviation, σ = 677
sample size, n = 100


Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96


ME = zc * σ/sqrt(n)
ME = 1.96 * 677/sqrt(100)
ME = 132.69

margin of error = 132.69

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3120 - 1.96 * 677/sqrt(100) , 3120 + 1.96 * 677/sqrt(100))
CI = (2987.31 , 3252.69)

(2987.31 < mu < 3252.69)


e)

Decrease because the sample size increased


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