Question

use the appropriate table in the Appendix of Tables to answer this question 11. [-/27 Points] DETAILS DEVORESTAT9 7.1.004. MY NOTES 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 o -2.2. (Round your answers to two decimal places.) (a) Computea 95% CI for when n. 25 and 2-52.0, watts (b) Computea 95% CI for when - 100 a 3- - 52.0 watts (c) Computea 99% CI for a when - 100 and 52.0 (d) Compute an 82% CI for w when n = 100 and 2-52.0. (e) How large 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) You may need to use the appropriate table in the Appendix of Tables to answer this question

Answer 1

The confidence interval is given by

= u52a/2 n

The critical values of z are

z = 1.96 for 95% CI

z = 2.58 for 99% CI

z = 1.34 for 82% CI

a) The 95% CI for $\mu$ when n = 25 and

T= 52.0

$\bg_white = 52.0 \pm 1.96 * \frac{2.2}{\sqrt{25}}$

$\bg_white = 52.0 \pm 0.8624$

b) The 95% CI for $\mu$ when n = 100 and

T= 52.0

$\bg_white = 52.0 \pm 1.96 * \frac{2.2}{\sqrt{100}}$

$\bg_white = 52.0 \pm 0.4312$

c) The 99% CI for $\mu$ when n = 100 and

T= 52.0

$\bg_white = 52.0 \pm 2.58 * \frac{2.2}{\sqrt{100}}$

$\bg_white = 52.0 \pm 0.5676$

d) The 82% CI for $\mu$ when n = 100 and

T= 52.0

$\bg_white = 52.0 \pm 1.34 * \frac{2.2}{\sqrt{100}}$

$\bg_white = 52.0 \pm 0.2948$

e) For the width to be equal to 1.0, the margin of error should be 0.5. The margin of error is given by

$ME = z_{\alpha/2} * \frac{\sigma}{\sqrt{n}}$

$0.5 = 2.58 * \frac{2.2}{\sqrt{n}}$

$\sqrt{n} = 2.58 * \frac{2.2}{0.5}$

Rounding up to the nearest integer, we have n = 129

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