Question

Two hospital emergency rooms use different procedures for triage of their patients. a local health care provider conducted a study to determine if there is a significant difference in the mean waiting time of patients for both hospitals. the 40 randomly selected subjects from medina general hospital (population 1) produce a mean waiting time of 18.3 minutes and a standard deviation of 2.1 minutes. the 50 randomly selected patients from southwest general hospital (population 2) produce a mean waiting time of 19.2 minutes and a standard deviation of 2.92 minutes. using a significance level of α = .02, what practical conclusion should be drawn from this study?

Answer 1

This is a difference of two means hypothesis test with the null and hypothesis tests as follows:


`H_0:\mu_1=\mu_2 \\H_A:\mu_1\\eq\mu_2`

The test statistics is given by


`z= \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{ \sqrt{ (\sigma_1^2)/(n_1)+ (\sigma_2^2)/(n_2) } }`

where:
`\bar{x}_1=18.3`

`\bar{x}_2=19.2`

`\mu_1-\mu_2=0`

`\sigma_1=2.1`

`n_1=40`

`\sigma_2=2.92`

`n_2=50`


`z= \frac{(18.3-19.2)-0}{ \sqrt{ ((2.1)^2)/(40)+ ((2.92)^2)/(50) } } \\ \\ = (-0.9)/( √(0.11025+0.170528) ) \\ \\ = (-0.9)/( √(0.280778) ) = (-0.9)/(0.5299) \\ \\ =-1.6985`

Since, the significance level is 0.02, the null hypothesis will be rejected if z < -2.33 or z > 2.33.

But, since z = -1.6985 which is not less than -2.33, we fail to reject the null hypothesis and we therefore, conclude that there is not enough evidence to conclude that there is a significant difference in the mean waiting time of patients for both hospitals.