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

Answer:

A sample size of 35 is needed.

Step-by-step explanation:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$W = z*\frac{\sigma}{\sqrt{n}}$

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 $\sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3$. So

$W = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.96*\frac{3}{\sqrt{n}}$

$\sqrt{n} = 1.96*3$

$(\sqrt{n})^2 = (1.96*3)^{2}$

$n = 34.57$

Rounding up

A sample size of 35 is needed.