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 1500 rpm. Assume that stray-load loss is normally distributed with sigma squared equals 9. How large must the sample size be if the width of the 95% interval for mu is to be 1.0

A sample size of 35 is needed.

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the width W as such

In which 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 , then . So

Rounding up

A sample size of 35 is needed.

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb