Question

A study conducted by a commuter train transportation authority involved surveying a random sample of 200 passengers. The results show that a customer had to wait on the average 9.3 minutes with a standard deviation of 6.2 minutes to buy his or her ticket. Find the upper confidence limit (UCL) for a 95% confidence interval for population mean , the true mean waiting time. Round your answer to two decimal places.

Answer 1

Answer:

The UCL of the 95% confidence interval for population mean is 10.16 minutes

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 $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

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.

$M = 1.96\frac{6.2}{\sqrt{200}} = 0.86$

The upper end of the interval is the sample mean added to M. So it is 9.3 + 0.86 = 10.16 minutes.

The UCL of the 95% confidence interval for population mean is 10.16 minutes