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Reserve Problems Chapter 11 Section 2 Problem 1 The department of health studied the number of patients who need liver transpCan you use gave a detailed instruction on how to input in excel ?

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Answer #1

For simplicity, let's assign t=0 to year 1999 and counting number of years after that. Then year 2000 corresponds to the value of t=1 and so on.

Put values for X, 't' and Y in different columns as shown -

X Y t
2000 1362 1
2001 1778 2
2002 3031 3
2003 4001 4
2004 5010 5
2005 7092 6
2006 7934 7
2007 7863 8
2008 8038 9
2009 8708 10
2010 8675 11
2011 8994 12
2012 9325 13

Then we need to perform simple linear regression for dependent variable 'Y' and independent variable 't'.

For this, you need to use the 'Data Analysis' add-on from the 'Data' tab to access 'Regression' function as shown -

File Home Insert Draw Page Layout Formulas Data Review View Help Power Pivot Share Le From Text/CSV DE Queries & Connections

Then a prompt box will open up asking you to input the dependent variable and independent variable. Fill in the corresponding ranges by just selecting the cell region for each 'Y' and 't'. Also make sure to mark the 'Labels' option as we have labels in our first row. Your input should be similar to this -

A B C D E F G H — J K T 1 x Y t 1362 1 N Regression ? 3 1778 3031 4001 2 3 Input Input Y Range: OK 1 4 $B$1:$B$14 4 5 6 7 Can

Thats it. You should get the regression output as shown -

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.948612089
R Square 0.899864896
Adjusted R Square 0.890761705
Standard Error 949.6871213
Observations 13
ANOVA
df SS MS F Significance F
Regression 1 89154801.78 89154801.78 98.85158599 7.83196E-07
Residual 11 9920961.912 901905.6284
Total 12 99075763.69
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 1393.846154 558.7469035 2.494593071 0.029791278 164.0525111 2623.639797
t 699.9010989 70.39549297 9.942413489 7.83196E-07 544.9616635 854.8405343

Thus, we can use this output to solve the question as shown-

(a)

The estimate of standard error of estimate is: Standard error = \widehat{\sigma} = 949.6871213

Thus, the estimate of \sigma^2 is: \widehat{\sigma}^2 = (949.6871213)^2 = 901905.6284

Rounding upto nearest integer gives: \widehat{\sigma}^2 = 901906

------------------------------------------------

(b)

The regression equation is written as -

\widehat{Y} = \widehat{\beta_{0}} + \widehat{\beta_{1}}t

Where \widehat{\beta_{1}} represents the rate of change of dependent variable 'Y' with respect to the independent variable 't'.

From the output, we get the slope coefficient as: 699.9010989

So, LTWL size changes by 700 in average for a year.

------------------------------------------

(c)

For year 2016, t = 2016 - 1999 = 17

And the regression equation is -

\widehat{Y} = 1393.846154 + (699.9010989)t

So, for t = 17 we get the predicted value of Y as -

\widehat{Y} = 1393.846154 + (699.9010989)(17) = 13292.16484

So, estimated number of patients in list = 13292

------------------------------------------------

(d)

For year 2011, t = 2011 - 1999 = 12

And the regression equation is -

\widehat{Y} = 1393.846154 + (699.9010989)t

So, for t = 12 we get the predicted value of Y as -

\widehat{Y} = 1393.846154 + (699.9010989)(12) = 9792.659314

So, estimated number of patients in list = 9793

And as the actual number of patients in list in 2011 = 8994 (from data given)

So, error = e = 9793 - 8994 = 799.

____________________________________________

Note that, you would get the same result if you run the regression using 'X' directly as the independent variable. You can try doing it for practice if you want. Its a general habit of parametrizing time in years to smaller scale in order to reduce the variance of data.

Please ask if you have any doubt(s) in comment section.

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