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

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Need help with answering the pdf file. It involves running R and knowing about regression


STAT 1124 HW2 Data : Anesthesiology Hrs.csv R Script : HW2 SRegr (Anesthesiology).R Distribution of Marks Question # 1 2 3 4 5 6 7 8 9 10 Total Marks 3 1 4 2 3 3 2 4 2 2 /26 Instructions: • Please hand in your HW on paper. All software output must copy-pasted if you type your HW, or printed if you handwrite your HW. • HW is due at the beginning of class on the due date. • You can do your HW in pairs. If you do so hand in only one HW with both your names. Hospital management have collected data over 24 months. For each month they record the total staff- hours (HOURS) needed to run the anaesthesiology department and the number of surgery cases (CASES) processed in that month. Load the data in RStudio and define the variables: y = HOURS x = CASES 1. Produce the summary statistics for the two variables. Fill the table below [3 marks]: n Min Q1 Q2 Mean Q3 Max SD r y = HOURS x = CASES 2. What constitutes an element (an observation) in these data? [1 mark] 3. Manually calculate the least-square (regression) line to predict HOURS using CASES. The value of r has been calculated to be 0.885. [4 marks] 4. Regress HOURS on CASES: (ie., use HOURS as the dependent variable, and CASES as the independent variable.) Show the R output for the summary of the regression model. (>summary(mod)), and the Scatter Plot. Copy/paste them to your HW, or print them if you handwrite the rest of your HW. (Compare the the results of your manual calculation above to the software output. They should be the same except for round-off error) [2 marks] 5. Identify the slope of the Regression Line and interpret what it means in the context of the problem. Use the number from software output. [3 marks] p. 1/2 6. Predict the anesthesiology staff time required (in hrs.) for a month when 400 surgeries are scheduled. Construct a prediction interval that will be correct 95% of the time. (Note, R calls the RMSE “Residual standard error.”) [3 marks] 7. Produce the R output for the Residual Plot and the Density Trace of the Residuals. [2 marks] 8. Are the two variables linearly related? Explain your answer by comparing r to the appropriate decision point, as well as by commenting on the overall shape shown in the Scatter/Residual Plot. [4 marks] 9. What does the Normality assumption mean in the context of this problem? [2 marks] 10. Is reasonable to assume that the Normality assumption has been satisfied in this problem? Justify your answer. [2 marks] p. 2/2
Answered Same DayMar 02, 2023

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

Subhanbasha answered on Mar 02 2023
48 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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