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A confidence interval is desired for the true average stray-load loss μ (measured in watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that the stray-load loss is normally distributed with σ = 3.0.

(a) Compute a 95% CI for μ when N = 25 and sample mean = 58.3.

(b) Compute a 95% CI for μ when N = 100 and sample mean = 58.3.

(c) Compute a 99% CI for μ when N = 100 and sample mean = 58.3.

(d) Compute a 82% CI for μ when N = 100 and sample mean = 58.3.

(e) How large must N be if the width of the 99% CI for μ is to be 1.0?

Expert Answer

Respuesta :

Answer:

Given below

Step-by-step explanation:

Given that population mean = 1500 and population std dev σ = 3.0.

Since sigma is known, we can use z critical values for finding out confidence intervals

a) 95% CI for μ when N = 25 and sample mean = 58.3

=[tex](58.3-1.96(\frac{3}{\sqrt{25} } ,58.3+1.96(\frac{3}{\sqrt{25} })\\= (58.3-1.176, 58.3+1.176)\\= (57.124, 59.476)[/tex]

b) a 95% CI for μ when N = 100 and sample mean = 58.3.

=[tex](58.3-1.96(\frac{3}{\sqrt{100} } ,58.3+1.96(\frac{3}{\sqrt{100} })\\= (58.3-0.588, 58.3+0.588)\\= (56.712, 58.588)\\[/tex]

c) a 99% CI for μ when N = 100 and sample mean = 58.3.

=[tex](58.3-2.58(\frac{3}{\sqrt{100} } ,58.3+2.58(\frac{3}{\sqrt{100} })\\= (58.3-0.774, 58.3+0.774)\\= (57.526, 59.074)[/tex]

d) a 82% CI for μ when N = 100 and sample mean = 58.3.

=[tex](58.3-1.33(\frac{3}{\sqrt{100} } ,58.3+1.33(\frac{3}{\sqrt{100} })\\= (58.3-0.399 58.3+0.399)\\= (57.9.1, 58.699)[/tex]

e) [tex]1=2.58(\frac{3}{\sqrt{n} } )\\\sqrt{n} =7.74\\n=59.9176[/tex]

n =60

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