A CI is desired for the true average stray-load loss (watts) for a certain type

Question

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 = 3.3. (Round your answers to two decimal places.)

(a) Compute a 95% CI for when n = 25 and x = 54.1.

( , )

watts

(b) Compute a 95% CI for when n = 100 and x = 54.1.

watts

(c) Compute a 99% CI for when n = 100 and x = 54.1.

watts

(d) Compute an 82% CI for when n = 100 and x = 54.1.

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.)
n =

Explanation / Answer

Solution:

(a)

Sample Mean = 54.1
SD = 3.3
Sample Size (n) = 25
Standard Error (SE) = SD/root(n) = 0.66
alpha (a) = 1 - 0.95 = 0.05

z critical value for 95% confidence interval:
z(a/2) = z(0.025) = 1.96

Margin of Error (ME) = z(a/2) x SE = 1.2936

95% confidence interval is given by:
Sample Mean +/- (Margin of Error) = 54.1 +/- 1.2936
= (52.8064 , 55.39)

b)

Sample Mean = 54.1
SD = 3.3
Sample Size (n) = 100
Standard Error (SE) = SD/root(n) = 0.33
alpha (a) = 1 - 0.95 = 0.05

z critical value for 95% confidence interval:
z(a/2) = z(0.025) = 1.96

Margin of Error (ME) = z(a/2) x SE = 0.6468

95% confidence interval is given by:
Sample Mean +/- (Margin of Error) = 54.1 +/- 0.6468
= (53.45 , 54.74)
c)

Sample Mean = 54.1
SD = 3.3
Sample Size (n) = 100
Standard Error (SE) = SD/root(n) = 0.33
alpha (a) = 1 - 0.99 = 0.01

z critical value for 99% confidence interval:
z(a/2) = z(0.005) = 2.5758

Margin of Error (ME) = z(a/2) x SE = 0.8500

99% confidence interval is given by:
Sample Mean +/- (Margin of Error) = 54.1 +/- 0.8500
= (53.25 , 54.95)
d)

Sample Mean = 54.1
SD = 3.3
Sample Size (n) = 100
Standard Error (SE) = SD/root(n) = 0.33
alpha (a) = 1 - 0.82 = 0.18

z critical value for 82% confidence interval:
z(a/2) = z(0.09) = 1.3408

Margin of Error (ME) = z(a/2) x SE = 0.4424

82% confidence interval is given by:
Sample Mean +/- (Margin of Error) = 51.4 +/- 0.4424
= (50.95 , 51.84)
e)

ME =0.5, need 99% CI, SD = 3.3
as ME = z*SE,
we want ME = 0.5, thus z*SD/root(n) = 0.5
This gives n = (SD * z /ME)^2 ,
where z = z_(alpha/2) = z_(0.005) = 2.5758
n = (3.3*2.5758/0.5)^2= 289.0095
Hence we should take n = 289

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