A hospital administrator hoping to improve wait times decides to estimate the average emergency room waiting time at her hospital. She collects a 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.
Q1: This confidence interval is not valid since we do not know if the population distribution of tthe ER wait times is nearly normal.
Ans: False. Reasoning Below.
1. Sample size is > 30 and sample is independent
2. Data is not abnormally skewed
3. With this we can say estimates of distribution will be normally distrubted by Central Limit Theorem
Q2: We are 95% confident that the average waiting time of these 64 emergency room patients is between 128 and 147 minutes?
Ans: False. We can be 100% certain about the mean for the sample
Q3: We are 95% confident that the average waiting time of all patients at this hospital’s ER is between 128 and 147 minutes.
Ans: True. This is what the confidence interval is used for. Using our sample, we are figuring the statistics of the total population.
Q4: 95% of random samples have a sample mean between 128 and 147 minutes.
Ans: False. This is a range around a sample mean, and the confidence interval is a measure of whether or not the population mean is within this range.
Q5: A 99% confidence interval would be narrower than the 95% confidence interval since we need to be more sure of our estimate.
Ans: False. The 99% confidence interval is actually wider.
Q5: The margin of error is 9.5 and the sample mean is 137.5.
Ans: We know z is 1.96 for a 95% confidence interval. We can solve for Standard Error using our interval.
Ans: Calculating the center of interval which is sample mean.:
sm <- mean(c(128,147))
sm## [1] 137.5SM is 137.5:
The margin of error moe is defined as:
\[ moe = z \times Standard Error \]
We know our confidence interval, so can solve for MOE below:
\[ C_{lower} = \bar{x} - 1.96 \times Standard Error \] \[ 128 = 137.5 - 1.96 \times Standard Error \]
Solving for moe, I got \(1.96 \times Standard Error = 9.5\). Hence we can conclude both the statements are true.
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
Ans: False.
\[ moe = Z \times Standard Error \]
\[ moe = Z \times \frac{\sigma_{sample}}{\sqrt{n}} \]