A CI is desired for the true average stray-load loss ? (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 stray-load loss is normally distributed with σ = 2.8. (Round your answers to two decimal places.)(a)Compute a 95% CI for ? when n = 25 and x = 51.4.(_________________, __________________) Watts(b) Compute a 95% CI for ? when n = 100 and x = 51.4.(________________________ , _____________________) watts(c) Compute a 99% CI for ? when n = 100 and x = 51.4.(___________________________, _______________________) watts(d) Compute an 82% CI for ? when n = 100 and x = 51.4.(_________________________, ___________________________) watts(e) How large must n be if the width of the 99% interval for ? is to be 1.0? (Round your answer up to the nearest whole number.)
Accepted Solution
A:
Answer:a) (50.30 , 52.50)b) (50.85 , 51.95)c) (50.68 , 52.12)d) (51.02 , 51.78)e) 209Step-by-step explanation:(a) Sample Mean = 51.4σ = 2.8Sample Size, n = 25
Standard Error, E = [tex]\frac{\sigma}{\sqrt{n}}[/tex] = 0.56z critical value for 95% confidence intervalz = 1.96Margin of Error (ME) = z × E = 1.0976
95% confidence interval is given as⇒ Mean ± ME = 51.4 ± 1.0976
or= (50.30 , 52.50)b) Sample Mean = 51.4σ = 2.8Sample Size, n = 100Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28z critical value for 95% confidence intervalz = 1.96Margin of Error (ME) = z × E = 0.5488
95% confidence interval is given as⇒ Mean ± ME = 51.4 ± 0.5488
or= (50.85 , 51.95)c) Sample Mean = 51.4σ = 2.8Sample Size, n = 100Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28z critical value for 99% confidence intervalz = 2.5758Margin of Error (ME) = z × E = 0.721299% confidence interval is given as⇒ Mean ± ME = 51.4 ± 0.7212
or= (50.68 , 52.12)d) Sample Mean = 51.4σ = 2.8Sample Size, n = 100Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28z critical value for 82% confidence intervalz = 1.3408Margin of Error (ME) = z × E = 0.375482% confidence interval is given as⇒ Mean ± ME = 51.4 ± 0.3754
or= (51.02 , 51.78)e) Margin of Error, ME = (width of interval) ÷ 2 = 0.5Now,σ = 2.8as ME = z × Standard Error,
z = 2.5758 for 99% confidence levelFor ME = 0.5, i,e [tex]\frac{z\times\sigma}{\sqrt{n}}[/tex] = 0.5
or[tex]\frac{2.5758
\times2.8}{\sqrt{n}}[/tex] = 0.5orn = [tex](\frac{2.5758
\times2.8}{0.5})^2[/tex] orn = 208.06 orn ≈ 209