Question

The average waiting time for a patient at an El Paso physician’s office is just over 29 minutes, well above the national average of 21 minutes. In fact, El Paso has the longest physician’s office waiting times in the United States. In order to address the issue of long patient wait times, some physician’s 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 Excel Online file below contains the data showing wait times (minutes) for a sample of patients at offices that do not have an office tracking system and wait times for a sample of patients at offices with an office tracking system. Construct a spreadsheet to answer the following questions.

a. What are the mean and median patient wait times for offices with a wait tracking system? What are the mean and median patient wait times for offices without a wait tracking system? Round your answers to 1 decimal place.

	Without Wait Tracking System	With Wait Tracking System
Mean
Median

b. What are the variance and standard deviation of patient wait times for offices with a wait tracking system? What are the variance and standard deviation of patient wait times for visits to offices without a wait tracking system? Please round the answers to 2 decimal place.

	Without Wait Tracking System	With Wait Tracking System
Variance
Standard deviation

c. Do offices with a wait tracking system have shorter patient wait times than offices without a wait tracking system?

(i) Offices with a wait tracking system have substantially longer patient wait times than offices without a wait tracking system.

(ii) Offices with a wait tracking system have substantially shorter patient wait times than offices without a wait tracking system.

(iii) Offices with a wait tracking system have about the same patient wait times as offices without a wait tracking system.

Choose the correct option.

_________Choice (i)Choice (ii)Choice (iii)

d. Considering only offices without a wait tracking system, what is the -score for the tenth patient in the sample (to 2 decimals)?

e. Considering only offices with a wait tracking system, what is the -score for the sixth patient in the sample (to 2 decimals)?

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

(i) The -score for the sixth patient in the sample who visited an office with a wait tracking system could not be compared with the -score calculated for part (d).

(ii) The -score for the sixth patient in the sample who visited an office with a wait tracking system is much larger only because that patient is part of a sample with a smaller mean.

(iii) The -score for the sixth patient in the sample who visited an office with a wait tracking system is much larger only because that patient is part of a sample with a smaller standard deviation.

(iv) The -score for the sixth patient in the sample who visited an office with a wait tracking system is much larger because that patient is part of a sample with a smaller mean and a smaller standard deviation.

Choose the correct answer.

_________Choice (i)Choice (ii)Choice (iii)Choice (iv)

f. Based on -scores, do the data for offices without a wait tracking system contain any outliers? Round your answers to 2 decimal places.

Patient's -scores	Without Wait Tracking System	With Wait Tracking System
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th

Based on -scores, do the data for offices with a wait tracking system contain any outliers?

Without Wait Tracking System	With Wait Tracking System			Without Wait Tracking System	With Wait Tracking System		Formulas for Column E	Formulas for Column F
24	31		Part a
67	11		Mean				#N/A	#N/A
17	13
20	19		Median				#N/A	#N/A
31	12
44	37		Part b
12	9		Variance				#N/A	#N/A
22	14
17	12		Standard Deviation				#N/A	#N/A
37	15
			Part d
			Z-score,10th patient				#N/A
			Part e
			Z-score, 6th patient					#N/A
			Part f
			1st Patient's Z-Score				#N/A	#N/A
			2nd Patient's Z-Score				#N/A	#N/A
			3rd Patient's Z-Score				#N/A	#N/A
			4th Patient's Z-Score				#N/A	#N/A
			5th Patient's Z-Score				#N/A	#N/A
			6th Patient's Z-Score				#N/A	#N/A
			7th Patient's Z-Score				#N/A	#N/A
			8th Patient's Z-Score				#N/A	#N/A
			9th Patient's Z-Score				#N/A	#N/A
			10th Patient's Z-Score				#N/A	#N/A

Answer 1

a)

Mean = Sum of Values /n

Median = Middle value after arranging the data in ascending order

	Without Wait Tracking System	With Wait Tracking System
Mean	29.1	17.3
Median	23	13.5

b)

Standard deviation of the two groups is calculated using the below formula:

Variance = Std Dev ²

	Without Wait Tracking System	With Wait Tracking System
Std Dev	16.56	9.30
Var	274.32	86.46

c)

Option ii ie Offices with a wait tracking system have substantially shorter patient wait times than offices without a wait tracking system

Given, n = 10 for both Groups

Let Group 1 be Without Wait Tracking System. Using the Mean and Std Dev values calculated in parts a & b.

alpha = 0.05

Null and Alternate Hypothesis

H₀: µ₁ = µ₂

H_a: µ₁ > µ₂

Test Statistic

Assuming, the population std deviation is not same.

t = (X_{1 ­­­}– X_{2 ­}– (µ₁ - µ₂))/ (s₁²/n₁ + s₂²/n₂ )^1/2 = 1.96

p-value = TDIST(1.96,10+10-2,1) = 0.033

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

Conclusion

Offices with a wait tracking system have shorter patient wait times than offices without a wait tracking system

d)

For 10^th patient in without a wait tracking system

X = 37

Z Score = (X-Mean)/ Std Dev = (37-29.1)/16.56 = 0.48

e)

For 6^th patient in with a wait tracking system

X = 37

Z Score = (X-Mean)/ Std Dev = (37-17.3)/9.30 = 2.12

Choice iv) ie The Z-score for the sixth patient in the sample who visited an office with a wait tracking system is much larger because that patient is part of a sample with a smaller mean and a smaller standard deviation

f)

Z Scores of Both Groups

Without Wait Tracking System	With Wait Tracking System
-0.31	1.47
2.29	-0.68
-0.73	-0.46
-0.55	0.18
0.11	-0.57
0.90	2.12
-1.03	-0.89
-0.43	-0.35
-0.73	-0.57
0.48	-0.25

On the basis on Z-scores, we conclude that the data for offices with a wait tracking system does not contain any outliers.