Question

Please Help with BOTH

1)

A CI is desired for the true average stray-load loss u (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 o = 2.6. (Round your answers to two decimal places.) (a) Compute a 95% CI for u when n = 25 and x = 53.0. 1 watts (b) Compute a 95% CI for u when n = 100 and x = 53.0. D ) watts (c) Compute a 99% CI for y when n = 100 and x = 53.0. D ) watts (d) Compute an 82% CI for u when n = 100 and 7 = 53.0. ) watts ( (e) How large must n be if the width of the 99% interval for u is to be 1.0? (Round your answer up to the nearest whole number.) You may need to use the appropriate table in the Appendix of Tables to answer this question.

2)

Answer 1

1) a) At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

t 20.025 /n

2.6 53 1.96 /25

53+1.02

51.98, 54.02

b) The 95% confidence interval is

t 20.025 /n

2.6 53 1.96 100

$= 53 \pm 0.51$

c) At 99% confidence level, the critical value is z_0.005 = 2.58

The 99% confidence interval is

$\bar x \pm z_{0.005} * \frac{\sigma}{\sqrt n}$

$= 53 \pm 2.58 * \frac{2.6}{\sqrt {100}}$

$= 53 \pm 0.67$

d) At 82% confidence level, the critical value is z_0.09 = 1.34

The 82% confidence interval is

$\bar x \pm z_{0.09} * \frac{\sigma}{\sqrt n}$

$= 53 \pm 1.34 * \frac{2.6}{\sqrt {100}}$

$= 53 \pm 0.35$

e) At 99% confidence level, the critical value is z_0.005 = 2.58

Margin of error = 1/2

$or, z_{0.005} * \frac{\sigma}{\sqrt n} = 0.5$

$or, 2.58 * \frac{2.6}{\sqrt n} = 0.5$

$or, n = (2.58 * \frac{2.6}{0.5})^2$

2)a) At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

t 20.025 /n

$= 4.85 \pm 1.96 * \frac{0.78}{\sqrt {22}}$

$= 4.85 \pm 0.33$

b) At 98% confidence level, the critical value is z_0.01 = 2.33

The 98% confidence interval is

$\bar x \pm z_{0.01} * \frac{\sigma}{\sqrt n}$

$= 4.56 \pm 2.58 * \frac{0.78}{\sqrt {16}}$

$= 4.56 \pm 0.45$

c) Margin of error = 0.47/2

$or, z_{0.025} * \frac{\sigma}{\sqrt n} = 0.235$

$or, 1.96 * \frac{2.6}{\sqrt n} = 0.235$

$or, n = (1.96 * \frac{2.6}{0.235})^2$

d) Margin of error = 0.22

$or, z_{0.005} * \frac{\sigma}{\sqrt n} = 0.22$

$or, 2.58 * \frac{2.6}{\sqrt n} = 0.22$

$or, n = (2.58 * \frac{2.6}{0.22})^2$