A Confidence interval is desired for the true average stray-load loss mu (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 15…

Question

asked Feb 5, 2021 138k views

1 Answer

Increddibelly · Answer 1 · 2021-02-09T19:17:42+0000

Answer:

A sample size of 35 is needed.

Explanation:

We have that to find our
$\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.95)/(2) = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of
$1-\alpha$ .

So it is z with a pvalue of
1-0.025 = 0.975 , so
z = 1.96

Now, find the width W as such

$W = z*(\sigma)/(√(n))$

In which
$\sigma$ is the standard deviation of the population and n is the size of the sample.

How large must the sample size be if the width of the 95% interval for mu is to be 1.0:

We need to find n for which W = 1.

We have that
$\sigma^(2) = 9$ , then
$\sigma = \sqrt{\sigma^(2)} = √(9) = 3$ . So

$W = z*(\sigma)/(√(n))$

1 = 1.96*(3)/(√(n))

√(n) = 1.96*3

(√(n))^2 = (1.96*3)^(2)

n = 34.57

Rounding up

A sample size of 35 is needed.