In: Math
Suppose that the average waiting time for a patient at a physician's office is just over 29 minutes. In order to address the issue of long patient wait times, some physicians' offices are using wait-tracking systems to notify patients of expected wait times. Patients can adjust their arrival times based on this information and spend less time in waiting rooms. The following data show wait times (minutes) for a sample of patients at offices that do not have a wait-tracking system and wait times for a sample of patients at offices with a wait-tracking system.
Without Wait- Tracking System |
With Wait-Tracking System |
24 | 14 |
60 | 9 |
10 | 33 |
22 | 10 |
34 | 18 |
47 | 30 |
12 | 12 |
12 | 17 |
27 | 9 |
30 | 3 |
(a) | Considering only offices without a wait-tracking system, what is the z-score for the 10th patient in the sample (wait time = 30 minutes)? |
If required, round your intermediate calculations and final answer to two decimal places. | |
z-score = | |
(b) | Considering only offices with a wait-tracking system, what is the z-score for the 6th patient in the sample (wait time = 30 minutes)? |
If required, round your intermediate calculations and final answer to two decimal places. | |
z-score = | |
How does this z-score compare with the z-score you calculated for part (a)? | |
(c) | Based on z-scores, do the data for offices without a wait-tracking system contain any outliers? |
Based on z-scores, do the data for offices with a wait-tracking system contain any outliers? | |
Solution:
In excel get the mean and standard deviation
Data>Data analysis>Descriptive statistics
we get
Without Wait-Tracking System | With-WaitTracking System | ||
Mean | 27.8 | Mean | 15.5 |
Standard Deviation | 16.03329868 | Standard Deviation | 9.489761 |
Sum | 278 | Sum | 155 |
Count | 10 | Count | 10 |
Solution(a)
z=x-mean/sd
z=30-27.8/16.03329868
z=0.14
z score=0.14
Solution(b)
z=x-mean/sd
z=30-15.5/9.489761
z=14.5/9.489761
z= 1.527963
z=1.53
z score=1.53
z score for with wait tracking system is high compared to z score for without wait tracking system.
Solution(b)
with z score we can find outliers using z scores
outliers <-3 or outlier above 3
Without Wait-Tracking System | mean | stddev | X-mean | Z-x-mean/sd |
24 | 27.8 | 16.0333 | -3.8 | -0.23701 |
60 | 27.8 | 16.0333 | 32.2 | 2.00832 |
10 | 27.8 | 16.0333 | -17.8 | -1.11019 |
22 | 27.8 | 16.0333 | -5.8 | -0.36175 |
34 | 27.8 | 16.0333 | 6.2 | 0.386695 |
47 | 27.8 | 16.0333 | 19.2 | 1.197508 |
12 | 27.8 | 16.0333 | -15.8 | -0.98545 |
12 | 27.8 | 16.0333 | -15.8 | -0.98545 |
27 | 27.8 | 16.0333 | -0.8 | -0.0499 |
30 | 27.8 | 16.0333 | 2.2 | 0.137214 |
With-WaitTracking System | Mean | Standard deviation | X-mean | z=x-mean/sd |
14 | 15.5 | 9.489761 | -1.5 | -0.15807 |
9 | 15.5 | 9.489761 | -6.5 | -0.68495 |
33 | 15.5 | 9.489761 | 17.5 | 1.844093 |
10 | 15.5 | 9.489761 | -5.5 | -0.57957 |
18 | 15.5 | 9.489761 | 2.5 | 0.263442 |
30 | 15.5 | 9.489761 | 14.5 | 1.527963 |
12 | 15.5 | 9.489761 | -3.5 | -0.36882 |
17 | 15.5 | 9.489761 | 1.5 | 0.158065 |
9 | 15.5 | 9.489761 | -6.5 | -0.68495 |
3 | 15.5 | 9.489761 | -12.5 | -1.31721 |
no outliers for offices without a wait-tracking system as no z value is below -3 or above 3
no outliers for offices with a wait-tracking system as no z value is below -3 or above 3