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 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.)

Answer 1

Answer:

a)  (50.30 , 52.50)

b) (50.85 , 51.95)

c) (50.68 , 52.12)

d)  (51.02 , 51.78)

e) 209

Step-by-step explanation:

(a)  Sample Mean = 51.4

σ = 2.8

Sample Size, n = 25

Standard Error, E = $\frac{\sigma}{\sqrt{n}}$ = 0.56

z critical value for 95% confidence interval

z = 1.96

Margin 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 = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = $\frac{2.8}{\sqrt{100}}$ = 0.28

z critical value for 95% confidence interval

z = 1.96

Margin 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 = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = $\frac{2.8}{\sqrt{100}}$ = 0.28

z critical value for 99% confidence interval

z = 2.5758

Margin of Error (ME) = z × E = 0.7212

99% confidence interval is given as

⇒ Mean ± ME

= 51.4 ± 0.7212

or

= (50.68 , 52.12)

d) Sample Mean = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = $\frac{2.8}{\sqrt{100}}$ = 0.28

z critical value for 82% confidence interval

z = 1.3408

Margin of Error (ME) = z × E = 0.3754

82% 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 = 0.5

Now,

σ = 2.8

as ME = z × Standard Error,

z = 2.5758  for 99% confidence level

For ME = 0.5,

i,e

$\frac{z\times\sigma}{\sqrt{n}}$ = 0.5

or

$\frac{2.5758 \times2.8}{\sqrt{n}}$ = 0.5

or

n = $(\frac{2.5758 \times2.8}{0.5})^2$

or

n = 208.06

or

n ≈ 209

Answer 2

The 95% confidence interval is  (50.30, 52.50).

How to calculate the confidence interval?

Sample Mean = 51.4

σ = 2.8

Sample Size, n = 25

Standard Error, E =  = 0.56

z critical value for 95% confidence interval, z = 1.96

Margin of Error (ME) = z × E = 1.0976

95% confidence interval is given as:

= Mean ± ME

= 51.4 ± 1.0976

= (50.30 , 52.50)

Sample Mean = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E =  = 0.28

z critical value for 95% confidence interval, z = 1.96

Margin of Error (ME) = z × E = 0.5488

95% confidence interval is given as:

= Mean ± ME

= 51.4 ± 0.5488

= (50.85 , 51.95)

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