Question

In: Statistics and Probability

Solution needed in excel file. A hospital conducted a study of the waiting time in its...

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

Solutions

Expert Solution

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.  


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