Question

A CI is desired for the true average stray-load loss μ (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 = 2.9. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 57.6.

(b) Compute a 95% CI for μ when n = 100 and x = 57.6.

(c) Compute a 99% CI for μ when n = 100 and x = 57.6.

(d) Compute an 82% CI for μ when n = 100 and x = 57.6.

(e) How large must n be if the width of the 99% interval for μ is to be 1.0? (Round your answer up to the nearest whole number.) n =

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

To compute the confidence intervals, use the formula CI = x ± Z * (σ / √n), where x is the sample mean, Z is the Z-value, σ is the standard deviation, and n is the sample size. For different confidence levels and sample sizes, the CI intervals can be calculated accordingly. In this case, the 95% CIs for different sample sizes and x values are computed. Additionally, the required sample size to achieve a specific CI width is determined.

Step-by-step explanation:

To compute the confidence intervals for the true average stray-load loss, we can use the formula:

CI = x ± Z * (σ / √n)

Where:

x = sample mean

Z = Z-value corresponding to the desired confidence level

σ = standard deviation

n = sample size

(a) For a 95% CI with n = 25 and x = 57.6:

Z-value for 95% confidence level = 1.96 (from standard normal distribution table)

CI = 57.6 ± (1.96 * (2.9 / √25))

CI ≈ 57.6 ± 1.143

CI ≈ (56.45, 58.75)

(b) For a 95% CI with n = 100 and x = 57.6:

Z-value for 95% confidence level = 1.96

CI = 57.6 ± (1.96 * (2.9 / √100))

CI ≈ 57.6 ± 0.569

CI ≈ (57.031, 58.169)

(c) For a 99% CI with n = 100 and x = 57.6:

Z-value for 99% confidence level = 2.58

CI = 57.6 ± (2.58 * (2.9 / √100))

CI ≈ 57.6 ± 0.746

CI ≈ (56.854, 58.346)

(d) For an 82% CI with n = 100 and x = 57.6:

Z-value for 82% confidence level = 1.386

CI = 57.6 ± (1.386 * (2.9 / √100))

CI ≈ 57.6 ± 0.448

CI ≈ (57.152, 58.048)

(e) To find the required sample size for a 99% CI with a width of 1.0:

Z-value for 99% confidence level = 2.58

CI width = 2 * (Z-value * (2.9 / √n))

1.0 = 2 * (2.58 * (2.9 / √n))

Solving for n:

1.0 / 5.16 = 2.9 / √n

√n = (2.9 * 5.16) / 1.0

n ≈ 26.505

Rounding up to the nearest whole number, n = 27