Question

A hospital conducted a study of the waiting time in its emergency room. The hospital has a main campus and three satellite locations. Management had a business objective of reducing waiting time for emergency room cases that did not require immediate attention. To study this, a random sample of 15 emergency room cases that did not require immediate attention at each location were selected on a particular day, and the waiting times (measured from check-in to when the patient was called into the clinic area) were collected.

Main	Satellite 1	Satellite 2	Satellite 3
120.08	30.75	75.86	54.05
81.90	61.83	37.88	38.82
78.79	26.40	68.73	36.85
63.83	53.84	51.08	32.83
79.77	72.30	50.21	52.94
47.94	53.09	58.47	34.13
79.88	27.67	86.29	69.37
48.63	52.46	62.90	78.52
55.43	10.64	44.84	55.95
64.06	53.50	64.17	49.61
64.99	37.28	50.68	66.40
53.82	34.31	47.97	76.06
62.43	66.00	60.57	11.37
65.07	8.99	58.37	83.51
81.02	29.75	30.40	39.17

1. At the 0.05 level of significance, is there evidence that the populations from which the samples are drawn are homogeneous (that is, no differences in variance)?
2. At the 0.05 level of significance, is there evidence of a difference in the mean waiting times in the four locations?
3. If appropriate, determine which locations differ in mean waiting time.

Answer 1

In order to test whether there were a differenec in the mean waiting times in the four locations at the 0.05 level of significance we apply ANOVA one factor model as follows:

We define the hypothesis as;

    

against      

In order to carry one way ANOVA we can follow this steps as follows:

Calculation of sum of square and total sum as follows;

Main		Satellite 1		Satellite 2		Satellite 3
120.08	14419.2064	30.75	945.5625	75.86	5754.74	54.05	2921.403
81.9	6707.61	61.83	3822.949	37.88	1434.894	38.82	1506.992
78.79	6207.8641	26.4	696.96	68.73	4723.813	36.85	1357.923
63.83	4074.2689	53.84	2898.746	51.08	2609.166	32.83	1077.809
79.77	6363.2529	72.3	5227.29	50.21	2521.044	52.94	2802.644
47.94	2298.2436	53.09	2818.548	58.47	3418.741	34.13	1164.857
79.88	6380.8144	27.67	765.6289	86.29	7445.964	69.37	4812.197
48.63	2364.8769	52.46	2752.052	62.9	3956.41	78.52	6165.39
55.43	3072.4849	10.64	113.2096	44.84	2010.626	55.95	3130.403
64.06	4103.6836	53.5	2862.25	64.17	4117.789	49.61	2461.152
64.99	4223.7001	37.28	1389.798	50.68	2568.462	66.4	4408.96
53.82	2896.5924	34.31	1177.176	47.97	2301.121	76.06	5785.124
62.43	3897.5049	66	4356	60.57	3668.725	11.37	129.2769
65.07	4234.1049	8.99	80.8201	58.37	3407.057	83.51	6973.92
81.02	6564.2404	29.75	885.0625	30.4	924.16	39.17	1534.289
1047.64	77808.4484	618.81	30792.05	848.42	50862.71	779.58	46232.34

Noe we define the sum of the values of the four samples as

Therefore the correction factor

Now the total sum of squares

Now sum of squares between the samples

with the degree of freedom

Now sum of squares within the samples

with the degree of freedom

Now ANOVA table can be written as;

Sources of variance	Sum of square	Degree of freedom	Mean square	Test statistics
Between sample	6312.444	3	2104.148
Within sample	18493.09	56	330.2338
Total	24805.53753

As the critical value of F statistics is 2.76 then the test criteria is

Therefore we accept the alternative hypthesis () and can conclude that the there were a differenec in the mean waiting times in the four locations at the 0.05 level of significance .

C)

To know which department has significant difference ,we can apply least significant difference test as

where MSE is the mean square error and r is the number of rows in each department.

Thus, if the absolute value of the difference between any two department means is greater than 82.64, we may conclude that they are not statistically equal.

Compare

Main and Satellite1:	428.83
Main and Satellite2:	199.22
Main and Satellite3:	268.06
Satellite 1and Satellite2:	-229.61
Satellite 1and Satellite3:	-160.77
Satellite 2and Satellite3:	68.84

Therefore only satellite 2 and satelite 3 has no mean difference but other have some statistically difference.

B) In order to test the difference between variance of the sample we define the hypothesis as;

    

against      

we apply the non parametric test .

If we run Excel dada analysis option:

Selecet four column input range and with alpha 0.05 , then OK.

Output is shown as follows;

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	15	1047.64	69.84267	331.3198
Column 2	15	618.81	41.254	375.976
Column 3	15	848.42	56.56133	205.3533
Column 4	15	779.58	51.972	408.2862
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	6312.444	3	2104.148	6.371691	0.000859	2.769431
Within Groups	18493.09	56	330.2338
Total	24805.54	59

Excell file