STAT 1124HW2Data : Anesthesiology Hrs.csvR Script : HW2 SRegr (Anesthesiology).RDistribution of MarksQuestion # XXXXXXXXXX10 TotalMarks XXXXXXXXXX /26Instructions:• Please hand in your...

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Answered Same DayMar 02, 2023

Answer To: STAT 1124HW2Data : Anesthesiology Hrs.csvR Script : HW2 SRegr (Anesthesiology).RDistribution...

Subhanbasha answered on Mar 02 2023
50 Votes
Answers
1).
Ans:
    
    n
    Min
    Q1
    Q2
    Mean
    Q3
    Max
    SD
    r
    y = HOURS
    24
    285
    1193
    2109.5
    2187.625
    3103.75
    4196
    1190.9
    0.8846
    x =
CASES
    24
    68
    218
    413.5
    463.4167
    734.25
    951
    288.3
    
2).
Ans:
Here from the above summary table the average time to run the anaesthesiology department is 2187.625 hours and the average cases in 463.4167. By observing the min and max columns from the above summary there is huge difference.
    The standard deviation also very high which means there is no standard form of the data means the time required in some months are very high compared to other months from the last two years of the data.
In this type of deviation data, we can use Q2 or median as the average metric. But still the average hours required is 2109.5 hours and cases are 413.5. which means for each case taking 5.10 hours of staff time.
By observing the correlation between cases and hours that is 0.8846 means strong positive correlation. If the cases are increased there is a chance of 88% to increase in the hours in the staff time.
3).
Ans:
Using the data, we can calculate the means and standard deviations as follows:
mean(x) = 463.4167
mean(y) = 2187.625
SDx = 288.3
SDy = 1190.9
Using the formula for the slope, we get:
b = r * (SDy / SDx) = 0.885 * (1190.9/ 288.3) = 3.655728
Using the formula for the intercept, we get:
a = mean(y) - b * mean(x) = 2187.625 - 3.655728 * 463.4167 = 493.4996
Therefore, the equation of the least-square line is:
y = 493.4996 + 3.655728x
4).
Ans:
Summary of the model
Call:
lm(formula...
SOLUTION.PDF

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