4.13 Waiting at an ER, Part I.

A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a simple random sample of 64 patients and determines the time (in minutes) between when they checked in to the ER until they were first seen by a doctor.

A 95% confidence interval based on this sample is (128 minutes, 147 minutes), which is based on the normal model for the mean.

Determine whether the following statements are true or false, and explain your reasoning.

A. This confidence interval is not valid since we do not know if the population distribution of the ER wait times is nearly Normal.

  • This is likely False. The sample size is greater than 30 and the sample is less than 10% of the population (all Emergency Room Visit Wait Times), so it is independent. We have no information which might suggest whether or not there is skew, so long as it is not overly skewed, we could consider this a reasonable confidence interval for mean of this sample.

B. We are 95% confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes.

  • False, this interval is an estimate of the population based not on the sample.

C. We are 95% confident that the average waiting time of all patients at this hospital’s emergency room is between 128 and 147 minutes.

  • True, the confidence interval (if properly estimated) should show us that 95% of the time, when a sample of 64 is randomly taken from the population, the true population mean should fall within this interval.

D. 95% of random samples have a sample mean between 128 and 147 minutes.

  • False, the confidence interval states how sure we can be that the population mean is within this interval, based on the sample estimates.

E. A 99% confidence interval would be narrower than the 95% confidence interval since we need to be more sure of our estimate.

  • False, this would actually be a wider interval, including MORE possible values of wait time to be sure that the mean was captured 99% of the time. To capture more estimates you need to have a wider range

F. The margin of error is 9.5 and the sample mean is 137.5.

  • True, the margin of error is the the amount added to and subtracted from the mean, the margin of error is of the mean is \[1.96*SE\] which is added to and substracted from the mean arrive at the interval. In this case we can check by subtracting 9.5 from 147, which gives us 137.5 and adding 9.5 to 127 aboth ways get us to 137.5, which is our sample mean

G. In order to decrease the margin of error of a 95% confidence interval to half of what it is now, we would need to double the sample size.

  • False, in order to cut the margin of error in half you would have to QUADRUPLE the sample sizes.
Calculate Standard deviation from the standard error formula

s/sqrt(n) where n=64
Z-score (1.65)
Margin of Error (9.5)

\[1.65*(s/sqrt(64))=9.5\] \[s/8=9.5/1.65\] \[s/8=5.757575\] \[s=8*5.757575\] \[s=46.06\]

Calcualte our new n size based on a margin of error of 4.5

\[1.65(46.06/sqrt(n))=4.5\] \[46.06/sqrt(n)=2.727272\] \[46.06 =2.727272(sqrt(n))\]
\[46.06/2.727272=sqrt(n)\]
\[16.8886 =sqrt(n)\]

n=((46.06*1.65)/4.5)^2

\[16.88^2 = n = 285.2\] The slight exageration in this value is due to rounding errors, when carried out with long decimals it comes out to 256 and some change.