My Notes OAsk Your Teach A is desired for the true average stray-load loss ? (wa

Question

My Notes OAsk Your Teach A is desired for the true average stray-load loss ? (watts) for a certain type of induction motor when the of 1500 rpm. Assume that stray-load loss is normally distributed with ?-3.1. (Round your answers to two decimal places.) line current is held at 10 amps for a speed (a) Compute a 95% ct for ? when n-25 and x-56.3. 16 020 55.02 , 57.50 x) watts (b) Compute a 95% CI forla when n-100 and x-56.3. 5508 ]X,5752 (c) Compute a 99% ci for ? when n-100,and x-56.3. (an- (d) Compute an 82% CI for ? when n " 100 and x-56.3. x watts 4.70 x ,157.90 ×)watts 5.4661X57.13 X watts must n be if the width of the 99% interval for is to be 1.07 (Round your answer up to the nearest whole number.) (e) How large n" appropriate table in the Appendix of Tables to answer this question Need Help?

Explanation / Answer

solution:-
(a) x = 56.3 , standard deviation = 3.1 , n = 25
95% confidence value is 1.96
confidence interval formula = x +/- z * s / sqrt(n)
=> 56.3 +/- 1.96 * 3.1/sqrt(25)
=> (55.08 , 57.52)

(b)x = 56.3 , standard deviation = 3.1 , n = 100
confidence interval is
56.3 +/- 1.96 * 3.1/sqrt(100)
=> (55.69 , 56.91)

(c) x = 56.3 , standard deviation = 3.1 , n = 100
99% confidence interval for z is 2.58
56.3 +/- 2.58 * 3.1/sqrt(100)
=> (55.50 , 57.10)

(d) x = 56.3 , standard deviation = 3.1 , n = 100
82% confidence value is 1.34
confidence interval is
56.3 +/- 1.34 * 3.1/sqrt(100)
=> (55.88 , 56.72)

(e) 99% confidence is 2.58
=> n = (2*2.58)(3.1/1.0)^2
=> n = 256

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