Question

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

Answer 1

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.