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

Answer:Given belowStep-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 intervalsa) 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