Question

wait times a hospital claims that 75% of people who come to its emergency room are seen by a doctor within 30 minutes of checking in. to verify this claim, an auditor inspects the medical records of 55 randomly selected patients who checked into the emergency room during the last year. only 32 (58.2%) of these patients were seen by a doctor within 30 minutes of checking in. if the wait time is less than 30 minutes for 75% of all patients in the emergency room, what is the probability that the proportion of patients who wait less than 30 minutes is 0.582 or less in a random sample of 55 patients? based on your answer to part (a), is there convincing evidence that less than 75% of all patients in the emergency room wait less than 30 minutes? explain your reasoning.

1. thienan
The observed proportion of 0.582 is statistically significantly lower than the claimed proportion of 0.75, which means there is no convincing evidence that less than 75% of all patients in the emergency room wait less than 30 minutes

### How to determine if there is convincing evidence that less than 75% of all patients in the emergency room wait less than 30 minutes?

To answer this question, we need to use a statistical test to determine the probability that we would observe a proportion of patients waiting less than 30 minutes as low as 0.582 or lower if the true proportion of patients waiting less than 30 minutes was really 75%.
To do this use a one-sample z-test for proportions. The null hypothesis for this test is that the true proportion of patients waiting less than 30 minutes is equal to the claimed proportion of 75%, while the alternative hypothesis is that the true proportion is less than 75%.
To conduct the test, first, you need to calculate the standard error of the sample proportion. The standard error is calculated as follows:
standard error = √((claimed proportion × (1 – claimed proportion)) / sample size)
Substituting the values from the problem, we get:
standard error =  √((0.75 × (1 – 0.75)) / 55) = 0.0584
Also, calculate the z-score for the observed sample proportion of 0.582:
z-score = (observed proportion – claimed proportion) / standard error
Substituting the values from the problem, we get:
z-score = (0.582 – 0.75) / 0.0584 = -2.88
Finally, we can use the z-score to find the probability of observing a proportion as low as 0.582 or lower if the true proportion was really 75%.
This probability is known as the p-value. A p-value less than 0.05 is often considered to be statistically significant and indicates that the observed difference is unlikely to be due to random chance.
To find the p-value, we can use a z-table to find the probability of observing a z-score as low as  -2.88 or lower. The p-value is 0.00398, which is less than 0.05.
Based on the p-value, there is convincing evidence that the true proportion of patients waiting less than 30 minutes in the emergency room is less than 75%.