In: Statistics and Probability
Solution needed in excel file.
A hospital conducted a study of the waiting time in its emergency room. The hospital has a main campus a three affiliated 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 and stores in ERWaiting (see folder). At the 0.05 level of significance, is there evidence of a difference in the mean waiting times in the four location?
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 |
Solution:
We can use the excel ANOVA: Single Factor data analysis tool to
find the answer to the given questions.
The excel steps are:
Enter the data in excel.
Click on Data > Data Analysis > ANOVA: Single Factor > OK
Input Range: Select the data range for all the data including labels
Mark Labels in the first row
Alpha = 0.05
Choose the output range and click OK.
The output is given below:
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Main | 15 | 1047.64 | 69.8427 | 331.3198 | ||
Satellite 1 | 15 | 618.81 | 41.2540 | 375.9760 | ||
Satellite 2 | 15 | 848.42 | 56.5613 | 205.3533 | ||
Satellite 3 | 15 | 779.58 | 51.9720 | 408.2862 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 6312.4439 | 3 | 2104.1480 | 6.3717 | 0.0009 | 2.7694 |
Within Groups | 18493.0937 | 56 | 330.2338 | |||
Total | 24805.5375 | 59 |
The null and alternative hypotheses are:
The test statistic is:
The p-value is:
Conclusion: Since the p-value is less than the significance level, we, therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that there is a difference in the mean waiting times in the four locations at the 0.05 significance level.