The hospital administrator mentioned in Exercise 4.13 randomly selected 64 patients and measured the time (in minutes) between when they checked in to the ER and the time they were first seen by a doctor. The average time is 137.5 minutes and the standard deviation is 39 minutes. She is getting grief from her supervisor on the basis that the wait times in the ER has increased greatly from last year’s average of 127 minutes. However, she claims that the increase is probably just due to chance.
(-) Independence: The 64 patients were randomly selected. It is reasonable to assume that this sample is less than 10% of all patients.
(-) Sample size: The sample size of 64 is greater than 30.
(-) Skewness: It is assumed that the data is not strongly skewed.
Based on the above points, the conditions for inference are met.
The null hypothesis is that the average wait times of ER patients have not changed from 127 minutes.
The alternative hypothesis is that the average wait times of ER patients is not 127 minutes.
\[H_0: \mu = 127\] \[H_A: \mu \ne 127\] Z-score:
\[Z = \frac{{\bar{x}} - H_0}{SE_{\bar{x}}} = \frac{137.5 - 127}{\frac{39}{\sqrt{64}}} = 2.15\]
\[p-value = P(Z < -2.15) + P(Z > 2.15) = 0.0158 + 0.0158 = 0.0316\]
0.0316 is less than the significance level of 0.05, therefore we reject the null hypothesis in favor of the alternative.
A p-value of 0.0316 is statistically significant evidence that the average wait time of ER patients has increased, using a significance level of 0.05.
If the significance level is changed to 0.01, the conclusion would change.
The p-value of 0.0316 is greater than 0.01, therefore we would fail to reject the null hypothesis and conclude that there is insufficient evidence that the average waiting of ER patients has changed.