Salaries of entry-level computer engineers have Normal distribution with unknown mean and variance. Three randomly selected computer engineers have salaries (in $ 1000s): (a) Construct a 90%...

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Given Data:
30, 50, 70
We calculate sample mean and std deviation from given data.
Sample Mean, x̄ =
∑
(x)
n
= 1503 = 50
Sample Variance, s2 =
∑
(x− x̄)2
n− 1 =
800
2 = 400
Sample std dev, s =
√
s2 =
√
400 = 20
A 90% CI for µ using t-dist
Sample Mean = x = 50
Sample Standard deviation = s = 20
Sample Size = n = 3
Significance level = α = 1− 0.9 = 0.1
Degrees of freedom for t-distribution, d.f . = n− 1 = 2
Critical Value = tα/2,df = t0.05,df=2 = 2.920 (from t-table, two-tails, d.f . = 2 )
Margin of Error = E = tα/2,df ×
sx√
n
= 2.920× 20√
3
E = 2.920× 11.547005
Margin of Error, E = 33.717255
Limits of 90% confidence interval are given by:
Lower limit = x− E = 50− 33.717255 ≈ 16.282745 ≈ 16.283
Upper limit = x+ E = 50 + 33.717255 ≈ 83.717255 ≈ 83.717
90% confidence interval is: x± E = 50± 33.717255
= (16.282745, 83.717255)
90% CI using t-dist: 16.283 < µ < 83.717
Note that exact answer using technology is:
(50 - 20/SQRT(3)*T.INV(1-(1-0.9)/2,3-1 ) ,
50 +...