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I have a regression model for some time series data investigating drug utilization. The purpose is to fit a spline to a time series and work out 95% CI etc. The model goes as follows:

id <- ts(1:length(drug$Date))

a1 <- ts(drug$Rate)

a2 <- lag(a1-1)

tg <- ts.union(a1,id,a2)

mg <-lm (a1~a2+bs(id,df=df1),data=tg) 

The summary output of mg is:

Call:

lm(formula = a1 ~ a2 + bs(id, df = df1), data = tg)

Residuals:

     Min       1Q   Median       3Q      Max 

-0.31617 -0.11711 -0.02897  0.12330  0.40442 

Coefficients:

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

(Intercept)        0.77443    0.09011   8.594 1.10e-11 ***

a2                 0.13270    0.13593   0.976  0.33329    

bs(id, df = df1)1 -0.16349    0.23431  -0.698  0.48832    

bs(id, df = df1)2  0.63013    0.19362   3.254  0.00196 ** 

bs(id, df = df1)3  0.33859    0.14399   2.351  0.02238 *  

---

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

I am using the Pr(>|t|) value of a2 to test if the data under investigation are autocorrelated.

Is it possible to extract this value of Pr(>|t|) (in this model 0.33329) and store it in a scalar to perform a logical test?

Alternatively, can it be worked out using another method?

1 Answer

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edited

The summary of the lm function is stored in a matrix called coefficient, so to access the required values, you can do the following:

a2Pval <- summary(mg)$coefficients[2, 4]

Or, more generally/

coef(summary(mg))["a2","Pr(>|t|)"]

If you want to explore more in R programming then watch this R programming tutorial for beginner:

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