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Problem 6 Simple Linear Regression To the Internal Revenue Service, the reasonableness of total itemized deductions depends on the taxpayers adjusted gross income. Large deductions, which include charity and medical deductions, are more reasonable for taxpayers with large adjusted gross incomes. If a taxpayer claims larger than average itemized deductions for a given level of income, the chances of an IRS audit are increased. Data (in thousands of dollars) on adjusted gross income and the average or reasonable amount of itemized deductions follow REASONABLE AMOUNT OF ITEMIZED DEDUCTIONS. ADJUSTED GROSS INCOME ($1000) 27 32 48 65 85 120 9.6 9.6 10.1 11.1 13.5 17.7 25.5 A computer output is produced to examine this relationship further: SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.977 0.955 0.946 1.372 7 Standard Lower Upper Loer Upper 95% CoefficientsError Stat P-value 95% 95.0% 95.0% Intercept 4.6771.033 4.526 0.006 2.020 7.333 2.020 7.333 Adjusted gross income 0.161 0.016 10.285 0.000 0.121 0.202 0.121 0.202 Answer the following questionsa. Develop an estimated regression equation that can be used to predict reasonable amount of itemized deductions. What is the coefficient of determination? Comment on the goodness of fit of the model What is the value of the sample correlation? Interpret this value. Interpret the slope coefficient Test whether the fitted regression model is statistically significant at the 5% level Use the estimated regression equation to predict reasonable amount of itemized deductions with an adjusted gross income of $52,500. b. c. d. e. f.

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Answer #1

(a)

Let Y : Amount of itemized deduction

X : Adjusted gross income

The required regression model is

y' = 4.677 + 0.161*x

(b)

The coefficient of determination is

R20,955

It shows that 95.5% of variation in Y is explained by X. Since R-square is large so it seems to be good fit.

(C)

The correlation coefficient is

r = 0.977

It is shows that relationship between the variables is strong.

(d)

The slope is 0.161. It shows that for each unit increase in X, y is increased by 0.161 units.

(e)

The p-value of slope is 0.000. Since p-value is less than 0.05 so model is significant.

(F)

The predicted value y for X = 52.5 is

y' = 4.677 + 0.161*52.5 = 13.1295

Answer: 13.13

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