# how do you determine the growth rate in from the linear model? [duplicate]

I have a time series data like this:

structure(list(TimePeriod = structure(c(1464926400, 1464927300,
1464928200, 1464929100, 1464930000, 1464930900, 1464931800, 1464932700,
1464933600, 1464934500), class = c("POSIXct", "POSIXt"), tzone = ""),
cpubusy = c(35.66, 37.05, 36.9, 36.66, 37.51, 37.2, 35.26,
36.81, 36.14, 36.18)), .Names = c("TimePeriod", "cpubusy"
), row.names = c(NA, 10L), class = "data.frame")


linear model:

lin<-lm(data=df, cpubusy~TimePeriod)


I am trying to read the output of lm function to determine the growth rate:

summary(lin)

Call:
lm(formula = cpubusy ~ TimePeriod, data = data1)

Residuals:
Min      1Q  Median      3Q     Max
-59.188 -17.771   0.182  18.622  86.633

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.661e+04  5.930e+02  -28.01   <2e-16 ***
TimePeriod   1.134e-05  4.041e-07   28.07   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 29.77 on 2514 degrees of freedom
Multiple R-squared:  0.2386,    Adjusted R-squared:  0.2383
F-statistic:   788 on 1 and 2514 DF,  p-value: < 2.2e-16


Would it be fair to assume that growth rate of cpubusy is equal to R-squared: 0.2386?

• No, the growth rate is estimated to be 1.134e-05. You should read some textbook on linear regression. – Roland Sep 1 '16 at 15:34
• @Roland, when I look at the linear line, the line is upward, with big slope. But, when I look at the coefficient (1.134e-05) is very low. How would I get the over growth rate? – user1471980 Sep 1 '16 at 15:43
• I'm not sure where you see a big slope in plot(cpubusy ~ TimePeriod, data = data1). – Roland Sep 1 '16 at 15:46
• @user1471980 try to read a bit more about how to interpret coefficients of a linear model. That 1.134e-05 is the increase of cpubusy for 1 unit increase in variable TimePeriod. So the interpretation depends on the units of that variable. Is it seconds maybe? Hours? Days? – AntoniosK Sep 1 '16 at 16:12
• This certainly looks like a very poor / poorly fitting model to me. – gung - Reinstate Monica Sep 1 '16 at 18:51

Check the following code to better understand my comment above. When you feed a date independent variable to a model you let R transform it to a number which messes up your interpretation. If you transform your variable to distance from first measurement (in minutes) you know exactly what you're trying to interpret.

structure(list(TimePeriod = structure(c(1464926400, 1464927300,
1464928200, 1464929100, 1464930000, 1464930900, 1464931800, 1464932700,
1464933600, 1464934500), class = c("POSIXct", "POSIXt"), tzone = ""),
cpubusy = c(35.66, 37.05, 36.9, 36.66, 37.51, 37.2, 35.26,
36.81, 36.14, 36.18)), .Names = c("TimePeriod", "cpubusy"
), row.names = c(NA, 10L), class = "data.frame") -> dt

dt$TimePeriod2 = seq(0,9)*15 dt$TimePeriod3 = as.numeric(dt\$TimePeriod)

dt

#             TimePeriod cpubusy TimePeriod2 TimePeriod3
# 1  2016-06-03 05:00:00   35.66           0  1464926400
# 2  2016-06-03 05:15:00   37.05          15  1464927300
# 3  2016-06-03 05:30:00   36.90          30  1464928200
# 4  2016-06-03 05:45:00   36.66          45  1464929100
# 5  2016-06-03 06:00:00   37.51          60  1464930000
# 6  2016-06-03 06:15:00   37.20          75  1464930900
# 7  2016-06-03 06:30:00   35.26          90  1464931800
# 8  2016-06-03 06:45:00   36.81         105  1464932700
# 9  2016-06-03 07:00:00   36.14         120  1464933600
# 10 2016-06-03 07:15:00   36.18         135  1464934500

lin<-lm(data=dt, cpubusy~TimePeriod)
summary(lin)

# Call:
#   lm(formula = cpubusy ~ TimePeriod, data = dt)
#
# Residuals:
#   Min      1Q  Median      3Q     Max
# -1.2166 -0.2359  0.1624  0.3733  0.9528
#
# Coefficients:
#   Estimate Std. Error t value Pr(>|t|)
# (Intercept)  6.564e+04  1.332e+05   0.493    0.635
# TimePeriod  -4.478e-05  9.095e-05  -0.492    0.636
#
# Residual standard error: 0.7435 on 8 degrees of freedom
# Multiple R-squared:  0.02941, Adjusted R-squared:  -0.09191
# F-statistic: 0.2424 on 1 and 8 DF,  p-value: 0.6357

lin<-lm(data=dt, cpubusy~TimePeriod2)
summary(lin)

# Call:
#   lm(formula = cpubusy ~ TimePeriod2, data = dt)
#
# Residuals:
#   Min      1Q  Median      3Q     Max
# -1.2166 -0.2359  0.1624  0.3733  0.9528
#
# Coefficients:
#   Estimate Std. Error t value Pr(>|t|)
# (Intercept) 36.718364   0.436968  84.030 4.49e-13 ***
#   TimePeriod2 -0.002687   0.005457  -0.492    0.636
# ---
#   Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 0.7435 on 8 degrees of freedom
# Multiple R-squared:  0.02941, Adjusted R-squared:  -0.09191
# F-statistic: 0.2424 on 1 and 8 DF,  p-value: 0.6357

lin<-lm(data=dt, cpubusy~TimePeriod3)
summary(lin)

# Call:
#   lm(formula = cpubusy ~ TimePeriod3, data = dt)
#
# Residuals:
#   Min      1Q  Median      3Q     Max
# -1.2166 -0.2359  0.1624  0.3733  0.9528
#
# Coefficients:
#   Estimate Std. Error t value Pr(>|t|)
# (Intercept)  6.564e+04  1.332e+05   0.493    0.635
# TimePeriod3 -4.478e-05  9.095e-05  -0.492    0.636
#
# Residual standard error: 0.7435 on 8 degrees of freedom
# Multiple R-squared:  0.02941, Adjusted R-squared:  -0.09191
# F-statistic: 0.2424 on 1 and 8 DF,  p-value: 0.6357


Model 1 and 3 are the same because that's what R does to your date variable in order to produce coefficients. 2nd model has same predictive capability, as expected, because you haven't changed your variable. You just transformed it to something more meaningful to you.

If you follow the approach of the 2nd model you know your variable is expressed in minutes. The coefficient obtained by the model, let's say C (positive), is your growth rate. And you can say that on average your dependent variable increases by C for every minute. Or C * 60 per hour, etc.

• @AntonioK, I need to calculate the slope of the linear model, that is the overall growth rate. – user1471980 Sep 1 '16 at 18:19