Question

A bank manager wishes to estimate the mean waiting time spent by customers at his bank. He knows from previous experience that the standard deviation is about 4.0 minutes. If he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be approximately 143.a.trueb.false

Answer 1

Answer:

b. False

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:

$\alpha = \frac{1-0.9}{2} = 0.05$

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 $1-0.05 = 0.95$, so $z = 1.645$

Now, find the margin of error M as such

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

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

If he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be

Sample size of n when $M = 1, \mu = 4$. So

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

$1 = 1.645*\frac{4}{\sqrt{n}}$

$\sqrt{n} = 4*1.645$

$(\sqrt{n})^{2} = (4*1.645)^{2}$

$n = 43.29$

Rouding up, 44.

Approximately 143.

False