Question

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 23 62 15 21 32 45 10 26 16 35 With Wait-Tracking System 31 10 13 17 35 10 16 (a) Considering only offices without a wait-tracking system, what is the z-score for the 10th patient in the sample (wait time 35 minutes)? If required, round your intermediate calculations and final answer to two decimal places. 2 score (b) Considering only offices with a wait-tracking system, what is the z-score for the 6th patient in the sample (wait time = 35 minutes)? If required, round your intermediate calculations and final answer to two decimal places. 2-score How does this z-score compare with the z-score you calculated for part (a)? The input in the box below will not be graded, but may be reviewed and considered by your instructor. blank

Answer 1

Let’s first calculate mean and standard deviation of given sample in following manner:

	Without Wait-	With Wait-Tracking
	Tracking System	System
	23	31
	62	10
	15	13
	21	17
	32	11
	45	35
	10	10
	26	2
	16	11
	35	16
Mean (µ) (Average)	28.50	15.60
Standard Deviation (σ)	15.71	10.07

1. Considering only offices without a wait-tracking system, what is the z-score for the 10th patient in the sample (wait time = 35 minutes)?

We know that

Z-score = (X – µ)/σ

Where,

X = 35 minutes

Mean µ = 28.50 minutes

Standard deviation σ = 15.71

Therefore, Z-score = (35 – 28.5)/15.71 = 0.4136

The z-score for the 10th patient in the sample is 0.41.    

2. Considering only offices with a wait-tracking system, what is the z-score for the 6th patient in the sample (wait time = 35 minutes)?

We know that

Z-score = (X – µ)/σ

Where,

X = 35 minutes

Mean µ = 15.60 minutes

Standard deviation σ = 10.07

Therefore, Z-score = (35 – 15.60)/10.07 = 1.93

The z-score for the 6th patient in the sample is 1.93.

How does this z-score compare with the z-score you calculated for part (a)?

Both z–scores are positive that means both patients had wait times more than the means of their respective samples. Both the patients have 35 minutes wait time but the z–score is smaller for the 10^th patient of part (a) in comparison of 6^th patient of part (b) because that patient is part of a sample with a higher mean and a higher standard deviation.

(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?

The z-score of all observations (calculated as above):

	Without Wait-	Z-score	With Wait-Tracking	Z-score
	Tracking System (X)	Z=(X – µ)/σ	System (X)	Z=(X – µ)/σ
	23	-0.35	31	1.53
	62	2.13	10	-0.56
	15	-0.86	13	-0.26
	21	-0.48	17	0.14
	32	0.22	11	-0.46
	45	1.05	35	1.93
	10	-1.18	10	-0.56
	26	-0.16	2	-1.35
	16	-0.80	11	-0.46
	35	0.41	16	0.04
Mean (µ)	28.50		15.60
Standard Deviation (σ)	15.71		10.07

Data for both without wait-tracking and with wait-tracking contains outliers as z-score is > 1 in many cases