Question

The data file wages contains monthly values of the average hourly wages (in dol-lars) for workers in the U.S. apparel and textile products industry for July 1981 through June 1987.

(a) Display and interpret the time series plot for these data.

(b) Use least squares to fit a linear time trend to this time series. Interpret the regression output. Save the standardized residuals from the fit for further analysis.

(c) Construct and interpret the time series plot of the standardized residuals from part (b).

(d) Use least squares to fit a quadratic time trend to the wages time series. Interpret the regression output. Save the standardized residuals from the fit for further analysis.

(e) Construct and interpret the time series plot of the standardized residuals from part (d).

data file wages

wages
7.75
7.74
7.87
7.89
7.94
8
8.17
8.1
8.13
8.19
8.22
8.25
8.31
8.26
8.33
8.31
8.36
8.42
8.46
8.5
8.47
8.48
8.49
8.49
8.53
8.44
8.52
8.54
8.61
8.67
8.72
8.7
8.73
8.75
8.76
8.77
8.83
8.78
8.85
8.87
8.93
9
9.07
9.07
9.09
9.13
9.13
9.13
9.19
9.12
9.16
9.17
9.23
9.31
9.31
9.33
9.33
9.33
9.34
9.32
9.35
9.27
9.31
9.32
9.37
9.42
9.44
9.44
9.44
9.48
9.45
9.44

Please use R to do it and provide the code

Answer 1

Solution :-

written using R

data(wages)

plot(wages, type='o', ylab='wages per hour')

#linear model

wages.lm = lm(wages~time(wages))

summary(wages.lm) #r square seems perfect

##

## Call:

## lm(formula = wages ~ time(wages))

##

## Residuals:

## Min 1Q Median 3Q Max

## -0.23828 -0.04981 0.01942 0.05845 0.13136

##

## Coefficients:

## Estimate Std. Error t value Pr(>|t|)   

## (Intercept) -5.490e+02 1.115e+01 -49.24 <2e-16 ***

## time(wages) 2.811e-01 5.618e-03 50.03 <2e-16 ***

## ---

## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

##

## Residual standard error: 0.08257 on 70 degrees of freedom

## Multiple R-squared: 0.9728, Adjusted R-squared: 0.9724

## F-statistic: 2503 on 1 and 70 DF, p-value: < 2.2e-16

plot(y=rstandard(wages.lm), x=as.vector(time(wages)), type = 'o')

#Quadratic model trend

wages.qm = lm(wages ~ time(wages) + I(time(wages)^2))

summary(wages.qm)

##

## Call:

## lm(formula = wages ~ time(wages) + I(time(wages)^2))

##

## Residuals:

## Min 1Q Median 3Q Max

## -0.148318 -0.041440 0.001563 0.050089 0.139839

##

## Coefficients:

## Estimate Std. Error t value Pr(>|t|)   

## (Intercept) -8.495e+04 1.019e+04 -8.336 4.87e-12 ***

## time(wages) 8.534e+01 1.027e+01 8.309 5.44e-12 ***

## I(time(wages)^2) -2.143e-02 2.588e-03 -8.282 6.10e-12 ***

## ---

## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

##

## Residual standard error: 0.05889 on 69 degrees of freedom

## Multiple R-squared: 0.9864, Adjusted R-squared: 0.986

## F-statistic: 2494 on 2 and 69 DF, p-value: < 2.2e-16

#time series plot of the standardized residuals

plot(y=rstandard(wages.qm), x=as.vector(time(wages)), type = 'o')