Step 1 :We are given that the national average waiting time for patients to be admitted into the hospital is 4 hours. We want to test if the average waiting time for patients at rural hospitals is less than this national average. We have a sample of 14 patients from rural hospitals with an average waiting time of 2.6 hours and a standard deviation of 2.3 hours. We will perform a hypothesis test at the 0.01 level of significance.
Step 2 :For this study, we should use a one-sample t-test because we are comparing the sample mean to a known population mean.
Step 3 :The null hypothesis is that the average waiting time for patients at rural hospitals is equal to or greater than the national average, \(H_{0}: \mu \geq 4\). The alternative hypothesis is that the average waiting time is less than the national average, \(H_{1}: \mu < 4\).
Step 4 :We calculate the test statistic using the formula for a one-sample t-test: \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\). Substituting the given values, we get \(t = \frac{2.6 - 4}{2.3/\sqrt{14}} \approx -2.278\).
Step 5 :We calculate the p-value using the t-distribution with \(n - 1 = 14 - 1 = 13\) degrees of freedom. The p-value is the probability of observing a test statistic as extreme as -2.278 or more extreme, given that the null hypothesis is true. Using a t-distribution table or calculator, we find that the p-value is approximately 0.0202.
Step 6 :We compare the p-value to the level of significance. Since the p-value (0.0202) is greater than the level of significance (0.01), we fail to reject the null hypothesis.
Step 7 :Thus, the final conclusion is that there is not enough evidence to conclude that the average waiting time for patients at rural hospitals is less than the national average of 4 hours.