lab 4 assignment
doug tharp
September 24, 2019
In the previous lab, we used the file statedata.csv to examine the relationship between life expectancy
and other socio-economic variables for the U.S. states. Using the same data set, now use stepwise
regression (step()) to obtain the best model to estimate life expectancy
setwd ("C:/doug/classes/geog6000/labs")
statedata <- read.csv("statedata.csv", row.names=1, na.strings="")
mycol <- heat.colors(5)
summary(statedata)## Population Income Illiteracy Life.Exp
## Min. : 365 Min. :3098 Min. :0.500 Min. :67.96
## 1st Qu.: 1080 1st Qu.:3993 1st Qu.:0.625 1st Qu.:70.12
## Median : 2838 Median :4519 Median :0.950 Median :70.67
## Mean : 4246 Mean :4436 Mean :1.170 Mean :70.88
## 3rd Qu.: 4968 3rd Qu.:4814 3rd Qu.:1.575 3rd Qu.:71.89
## Max. :21198 Max. :6315 Max. :2.800 Max. :73.60
## Murder HS.Grad Frost Area
## Min. : 1.400 Min. :37.80 Min. : 0.00 Min. : 1049
## 1st Qu.: 4.350 1st Qu.:48.05 1st Qu.: 66.25 1st Qu.: 36985
## Median : 6.850 Median :53.25 Median :114.50 Median : 54277
## Mean : 7.378 Mean :53.11 Mean :104.46 Mean : 70736
## 3rd Qu.:10.675 3rd Qu.:59.15 3rd Qu.:139.75 3rd Qu.: 81163
## Max. :15.100 Max. :67.30 Max. :188.00 Max. :566432You will first need to make a
new dataframe that excludes the first column of statedata (statedata2 <- statedata[,-1]).
statedata2 <- statedata[,-1]• Build the null model (mod0) and full model (mod1)
nullmodel <- lm(Life.Exp ~ 1, data = statedata2)
summary(nullmodel)##
## Call:
## lm(formula = Life.Exp ~ 1, data = statedata2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.9186 -0.7611 -0.2036 1.0139 2.7214
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 70.8786 0.1898 373.4 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.342 on 49 degrees of freedomfullmodel <- lm(Life.Exp ~ ., data = statedata2)
summary(fullmodel)##
## Call:
## lm(formula = Life.Exp ~ ., data = statedata2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.39934 -0.53722 0.08628 0.53270 1.28452
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.112e+01 1.788e+00 39.771 < 2e-16 ***
## Income 1.219e-04 2.363e-04 0.516 0.6088
## Illiteracy -1.399e-01 3.617e-01 -0.387 0.7008
## Murder -2.742e-01 4.517e-02 -6.070 2.89e-07 ***
## HS.Grad 4.155e-02 2.352e-02 1.767 0.0843 .
## Frost -7.495e-03 3.056e-03 -2.452 0.0183 *
## Area -2.625e-07 1.706e-06 -0.154 0.8784
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7632 on 43 degrees of freedom
## Multiple R-squared: 0.7164, Adjusted R-squared: 0.6768
## F-statistic: 18.1 on 6 and 43 DF, p-value: 2.41e-10• Use the step() function to perform automatic stepwise model building. Report the R code you used
stepmodelfit <- step(nullmodel, scope=formula(fullmodel))## Start: AIC=30.44
## Life.Exp ~ 1
##
## Df Sum of Sq RSS AIC
## + Murder 1 53.838 34.461 -14.609
## + Illiteracy 1 30.578 57.721 11.179
## + HS.Grad 1 29.931 58.368 11.737
## + Income 1 10.223 78.076 26.283
## + Frost 1 6.064 82.235 28.878
## <none> 88.299 30.435
## + Area 1 1.017 87.282 31.856
##
## Step: AIC=-14.61
## Life.Exp ~ Murder
##
## Df Sum of Sq RSS AIC
## + HS.Grad 1 4.691 29.770 -19.925
## + Frost 1 3.135 31.327 -17.378
## + Income 1 2.405 32.057 -16.226
## <none> 34.461 -14.609
## + Area 1 0.470 33.992 -13.295
## + Illiteracy 1 0.273 34.188 -13.007
## - Murder 1 53.838 88.299 30.435
##
## Step: AIC=-19.93
## Life.Exp ~ Murder + HS.Grad
##
## Df Sum of Sq RSS AIC
## + Frost 1 4.3987 25.372 -25.920
## <none> 29.770 -19.925
## + Illiteracy 1 0.4419 29.328 -18.673
## + Area 1 0.2775 29.493 -18.394
## + Income 1 0.1022 29.668 -18.097
## - HS.Grad 1 4.6910 34.461 -14.609
## - Murder 1 28.5974 58.368 11.737
##
## Step: AIC=-25.92
## Life.Exp ~ Murder + HS.Grad + Frost
##
## Df Sum of Sq RSS AIC
## <none> 25.372 -25.920
## + Income 1 0.182 25.189 -24.280
## + Illiteracy 1 0.172 25.200 -24.259
## + Area 1 0.026 25.346 -23.970
## - Frost 1 4.399 29.770 -19.925
## - HS.Grad 1 5.955 31.327 -17.378
## - Murder 1 32.756 58.128 13.531The model steps forward through the variables bizarrely reducing the AIC further and fruther into negative territory. I’ll have to look up what the heck a negative AIC can even mean. the final AIC is a negative 25.92, six below the prior model without frost. It then consists of murder rate, frost and high school grad rate. Note that many demographers often include a dummy for southern state when modeling by US states as they perform worse on almost all desireable variables (here life expectancy). I assume frost is an indicator of sunbelt states which dilutes southern (the traditional american south) presumably adding in Arizona, etc. it appears that the data might be the number of days of frost per year. How that is weighted by city, population, mountains versus desert is unknown.
finalstepfit <- lm(Life.Exp ~ Frost + HS.Grad + Murder, data = statedata2)• Describe the final model obtained, together with its AIC score
summary(finalstepfit)##
## Call:
## lm(formula = Life.Exp ~ Frost + HS.Grad + Murder, data = statedata2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.5015 -0.5391 0.1014 0.5921 1.2268
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 71.036379 0.983262 72.246 < 2e-16 ***
## Frost -0.006912 0.002447 -2.824 0.00699 **
## HS.Grad 0.049949 0.015201 3.286 0.00195 **
## Murder -0.283065 0.036731 -7.706 8.04e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7427 on 46 degrees of freedom
## Multiple R-squared: 0.7127, Adjusted R-squared: 0.6939
## F-statistic: 38.03 on 3 and 46 DF, p-value: 1.634e-12finalstepfit <- lm(Life.Exp ~ Frost + HS.Grad + Murder, data = statedata2)• From the summary() function, report the goodness-of-fit (F-statistic and associated p-value) and the R2
The frost variable has a negative sign which is the opposite of what I would have expected, given the above discussion. However, it is possible that since many retirees move to southern states, a prepodnerance of the very old actually have as their final residence an address in AZ, FL or TX. The F stat of 38.03 is highly significant with an astoundnig p value of 0.000000000001634. the adjusted R2 is 0.69, suggesting a relatively tight fit on so few obs. These results do not match my expectations.
• Using the final model you obtain, make a prediction for the life expectancy in Utah in 2009,
Assuming the data are from 2008, then without adjustment I would build the model as:
print (statedata2, which = "UT" )## Income Illiteracy Life.Exp Murder HS.Grad Frost Area
## AL 3624 2.1 69.05 15.1 41.3 20 50708
## AK 6315 1.5 69.31 11.3 66.7 152 566432
## AZ 4530 1.8 70.55 7.8 58.1 15 113417
## AR 3378 1.9 70.66 10.1 39.9 65 51945
## CA 5114 1.1 71.71 10.3 62.6 20 156361
## CO 4884 0.7 72.06 6.8 63.9 166 103766
## CT 5348 1.1 72.48 3.1 56.0 139 4862
## DE 4809 0.9 70.06 6.2 54.6 103 1982
## FL 4815 1.3 70.66 10.7 52.6 11 54090
## GA 4091 2.0 68.54 13.9 40.6 60 58073
## HI 4963 1.9 73.60 6.2 61.9 0 6425
## ID 4119 0.6 71.87 5.3 59.5 126 82677
## IL 5107 0.9 70.14 10.3 52.6 127 55748
## IN 4458 0.7 70.88 7.1 52.9 122 36097
## IA 4628 0.5 72.56 2.3 59.0 140 55941
## KS 4669 0.6 72.58 4.5 59.9 114 81787
## KY 3712 1.6 70.10 10.6 38.5 95 39650
## LA 3545 2.8 68.76 13.2 42.2 12 44930
## ME 3694 0.7 70.39 2.7 54.7 161 30920
## MD 5299 0.9 70.22 8.5 52.3 101 9891
## MA 4755 1.1 71.83 3.3 58.5 103 7826
## MI 4751 0.9 70.63 11.1 52.8 125 56817
## MN 4675 0.6 72.96 2.3 57.6 160 79289
## MS 3098 2.4 68.09 12.5 41.0 50 47296
## MO 4254 0.8 70.69 9.3 48.8 108 68995
## MT 4347 0.6 70.56 5.0 59.2 155 145587
## NE 4508 0.6 72.60 2.9 59.3 139 76483
## NV 5149 0.5 69.03 11.5 65.2 188 109889
## NH 4281 0.7 71.23 3.3 57.6 174 9027
## NJ 5237 1.1 70.93 5.2 52.5 115 7521
## NM 3601 2.2 70.32 9.7 55.2 120 121412
## NY 4903 1.4 70.55 10.9 52.7 82 47831
## NC 3875 1.8 69.21 11.1 38.5 80 48798
## ND 5087 0.8 72.78 1.4 50.3 186 69273
## OH 4561 0.8 70.82 7.4 53.2 124 40975
## OK 3983 1.1 71.42 6.4 51.6 82 68782
## OR 4660 0.6 72.13 4.2 60.0 44 96184
## PA 4449 1.0 70.43 6.1 50.2 126 44966
## RI 4558 1.3 71.90 2.4 46.4 127 1049
## SC 3635 2.3 67.96 11.6 37.8 65 30225
## SD 4167 0.5 72.08 1.7 53.3 172 75955
## TN 3821 1.7 70.11 11.0 41.8 70 41328
## TX 4188 2.2 70.90 12.2 47.4 35 262134
## UT 4022 0.6 72.90 4.5 67.3 137 82096
## VT 3907 0.6 71.64 5.5 57.1 168 9267
## VA 4701 1.4 70.08 9.5 47.8 85 39780
## WA 4864 0.6 71.72 4.3 63.5 32 66570
## WV 3617 1.4 69.48 6.7 41.6 100 24070
## WI 4468 0.7 72.48 3.0 54.5 149 54464
## WY 4566 0.6 70.29 6.9 62.9 173 97203utah = 71.036379 + 4.5 * -0.283065 + 67.3 * 0.049949 + 137 * -0.006912
print (utah)## [1] 72.17721# 72.17721alternatively, using the predict function
predict(finalstepfit, level = 0.95, interval = "conf", statedata2[44,]) ## fit lwr upr
## UT 72.17723 71.74608 72.60837plot(finalstepfit)
# more stuff out of curiousity
plot (Life.Exp ~ log(Population), data = statedata, main = "life exp x log population, red = full model derived abline, blue = nullmodel")
abline(fullmodel, col = "red") ## Warning in abline(fullmodel, col = "red"): only using the first two of 7
## regression coefficientsabline(nullmodel, col = "blue", h=5, lty = 3)
given an increase in population size to approximately 2 785 000 in 2009, increase in high school
graduation (known) to 75% and a change in murder rate to 1.3/100 000. To do this you will
need to make a new dataframe. This can be done easily by selecting the appropriate row from
statedata (newstate <- statedata[44,]), then adjusting the values directly (newstate[2] <- 2785).
Note that in following the instructions for the first part of exercise 1, I dropped the first variable, which was population. so, here I will begin again, leaving it in.
summary(statedata)## Population Income Illiteracy Life.Exp
## Min. : 365 Min. :3098 Min. :0.500 Min. :67.96
## 1st Qu.: 1080 1st Qu.:3993 1st Qu.:0.625 1st Qu.:70.12
## Median : 2838 Median :4519 Median :0.950 Median :70.67
## Mean : 4246 Mean :4436 Mean :1.170 Mean :70.88
## 3rd Qu.: 4968 3rd Qu.:4814 3rd Qu.:1.575 3rd Qu.:71.89
## Max. :21198 Max. :6315 Max. :2.800 Max. :73.60
## Murder HS.Grad Frost Area
## Min. : 1.400 Min. :37.80 Min. : 0.00 Min. : 1049
## 1st Qu.: 4.350 1st Qu.:48.05 1st Qu.: 66.25 1st Qu.: 36985
## Median : 6.850 Median :53.25 Median :114.50 Median : 54277
## Mean : 7.378 Mean :53.11 Mean :104.46 Mean : 70736
## 3rd Qu.:10.675 3rd Qu.:59.15 3rd Qu.:139.75 3rd Qu.: 81163
## Max. :15.100 Max. :67.30 Max. :188.00 Max. :566432brewers_desired_nullmodel <- lm(Life.Exp ~1, data = statedata)
brewers_desired_fullmodel <- lm(Life.Exp ~., data = statedata)
brewers_stepmodel <- step(brewers_desired_nullmodel, scope = formula(brewers_desired_fullmodel))## Start: AIC=30.44
## Life.Exp ~ 1
##
## Df Sum of Sq RSS AIC
## + Murder 1 53.838 34.461 -14.609
## + Illiteracy 1 30.578 57.721 11.179
## + HS.Grad 1 29.931 58.368 11.737
## + Income 1 10.223 78.076 26.283
## + Frost 1 6.064 82.235 28.878
## <none> 88.299 30.435
## + Area 1 1.017 87.282 31.856
## + Population 1 0.409 87.890 32.203
##
## Step: AIC=-14.61
## Life.Exp ~ Murder
##
## Df Sum of Sq RSS AIC
## + HS.Grad 1 4.691 29.770 -19.925
## + Population 1 4.016 30.445 -18.805
## + Frost 1 3.135 31.327 -17.378
## + Income 1 2.405 32.057 -16.226
## <none> 34.461 -14.609
## + Area 1 0.470 33.992 -13.295
## + Illiteracy 1 0.273 34.188 -13.007
## - Murder 1 53.838 88.299 30.435
##
## Step: AIC=-19.93
## Life.Exp ~ Murder + HS.Grad
##
## Df Sum of Sq RSS AIC
## + Frost 1 4.3987 25.372 -25.920
## + Population 1 3.3405 26.430 -23.877
## <none> 29.770 -19.925
## + Illiteracy 1 0.4419 29.328 -18.673
## + Area 1 0.2775 29.493 -18.394
## + Income 1 0.1022 29.668 -18.097
## - HS.Grad 1 4.6910 34.461 -14.609
## - Murder 1 28.5974 58.368 11.737
##
## Step: AIC=-25.92
## Life.Exp ~ Murder + HS.Grad + Frost
##
## Df Sum of Sq RSS AIC
## + Population 1 2.064 23.308 -28.161
## <none> 25.372 -25.920
## + Income 1 0.182 25.189 -24.280
## + Illiteracy 1 0.172 25.200 -24.259
## + Area 1 0.026 25.346 -23.970
## - Frost 1 4.399 29.770 -19.925
## - HS.Grad 1 5.955 31.327 -17.378
## - Murder 1 32.756 58.128 13.531
##
## Step: AIC=-28.16
## Life.Exp ~ Murder + HS.Grad + Frost + Population
##
## Df Sum of Sq RSS AIC
## <none> 23.308 -28.161
## + Income 1 0.006 23.302 -26.174
## + Illiteracy 1 0.004 23.304 -26.170
## + Area 1 0.001 23.307 -26.163
## - Population 1 2.064 25.372 -25.920
## - Frost 1 3.122 26.430 -23.877
## - HS.Grad 1 5.112 28.420 -20.246
## - Murder 1 34.816 58.124 15.528summary(brewers_stepmodel)##
## Call:
## lm(formula = Life.Exp ~ Murder + HS.Grad + Frost + Population,
## data = statedata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.47095 -0.53464 -0.03701 0.57621 1.50683
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.103e+01 9.529e-01 74.542 < 2e-16 ***
## Murder -3.001e-01 3.661e-02 -8.199 1.77e-10 ***
## HS.Grad 4.658e-02 1.483e-02 3.142 0.00297 **
## Frost -5.943e-03 2.421e-03 -2.455 0.01802 *
## Population 5.014e-05 2.512e-05 1.996 0.05201 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7197 on 45 degrees of freedom
## Multiple R-squared: 0.736, Adjusted R-squared: 0.7126
## F-statistic: 31.37 on 4 and 45 DF, p-value: 1.696e-12shockingly, these results are nearly identical. However, my shock is dulled by noticing that the coefficient on population is not significant at the traditional 0.05 level
predict(brewers_stepmodel, level = 0.95, interval = "conf", statedata[44,]) ## fit lwr upr
## UT 72.05753 71.62238 72.49269plot(brewers_stepmodel, main = "Brewers Desired Step Model")
newstate <- statedata[44,]then adjusting the values directly
• Use the step() function to perform automatic stepwise model building. Report the R code you used
Give the new expectancy plus 95% confidence interval
summary(newstate)## Population Income Illiteracy Life.Exp Murder
## Min. :1203 Min. :4022 Min. :0.6 Min. :72.9 Min. :4.5
## 1st Qu.:1203 1st Qu.:4022 1st Qu.:0.6 1st Qu.:72.9 1st Qu.:4.5
## Median :1203 Median :4022 Median :0.6 Median :72.9 Median :4.5
## Mean :1203 Mean :4022 Mean :0.6 Mean :72.9 Mean :4.5
## 3rd Qu.:1203 3rd Qu.:4022 3rd Qu.:0.6 3rd Qu.:72.9 3rd Qu.:4.5
## Max. :1203 Max. :4022 Max. :0.6 Max. :72.9 Max. :4.5
## HS.Grad Frost Area
## Min. :67.3 Min. :137 Min. :82096
## 1st Qu.:67.3 1st Qu.:137 1st Qu.:82096
## Median :67.3 Median :137 Median :82096
## Mean :67.3 Mean :137 Mean :82096
## 3rd Qu.:67.3 3rd Qu.:137 3rd Qu.:82096
## Max. :67.3 Max. :137 Max. :82096newstate[1] <- 2785
newstate[6] <- 75
newstate[5] <- 1.3
summary(newstate)## Population Income Illiteracy Life.Exp Murder
## Min. :2785 Min. :4022 Min. :0.6 Min. :72.9 Min. :1.3
## 1st Qu.:2785 1st Qu.:4022 1st Qu.:0.6 1st Qu.:72.9 1st Qu.:1.3
## Median :2785 Median :4022 Median :0.6 Median :72.9 Median :1.3
## Mean :2785 Mean :4022 Mean :0.6 Mean :72.9 Mean :1.3
## 3rd Qu.:2785 3rd Qu.:4022 3rd Qu.:0.6 3rd Qu.:72.9 3rd Qu.:1.3
## Max. :2785 Max. :4022 Max. :0.6 Max. :72.9 Max. :1.3
## HS.Grad Frost Area
## Min. :75 Min. :137 Min. :82096
## 1st Qu.:75 1st Qu.:137 1st Qu.:82096
## Median :75 Median :137 Median :82096
## Mean :75 Mean :137 Mean :82096
## 3rd Qu.:75 3rd Qu.:137 3rd Qu.:82096
## Max. :75 Max. :137 Max. :82096UTpredict <- predict(brewers_stepmodel, level = 0.95, interval = "conf", newstate[1,])
summary(UTpredict)## fit lwr upr
## Min. :73.46 Min. :72.84 Min. :74.07
## 1st Qu.:73.46 1st Qu.:72.84 1st Qu.:74.07
## Median :73.46 Median :72.84 Median :74.07
## Mean :73.46 Mean :72.84 Mean :74.07
## 3rd Qu.:73.46 3rd Qu.:72.84 3rd Qu.:74.07
## Max. :73.46 Max. :72.84 Max. :74.07• Do the same for California. 2009 population = 36 962 000; 2009 high school graduation = 68.3%;
2009 murder rate = 5.3/100 000 (figures are approximate)
print(statedata, which = “CA”) #I counted down to find out which was CA, I’ll learn eventually how to do a which statement
statedata5 <- statedata[5,]
statedata5[1] <- 36962
statedata5[6] <- 68.3
statedata5[5] <- 5.3
CApredict <- predict(brewers_stepmodel, level = 0.95, interval = "conf", statedata5[1,])
summary(CApredict)## fit lwr upr
## Min. :74.35 Min. :72.65 Min. :76.06
## 1st Qu.:74.35 1st Qu.:72.65 1st Qu.:76.06
## Median :74.35 Median :72.65 Median :76.06
## Mean :74.35 Mean :72.65 Mean :76.06
## 3rd Qu.:74.35 3rd Qu.:72.65 3rd Qu.:76.06
## Max. :74.35 Max. :72.65 Max. :76.062. The file normtemp.csv contains measurements of body temperature from 130 healthy human subjects,
as well as their weight and sex. Use this data to model body temperatures as a function of both weight
and sex. You will need to convert the sex variable into a factor in order for R to recognize this as
a dummy variable (bodytemp\(sex <- factor(bodytemp\)sex, labels=c(“male”,“female”))). You
should also center the weight variable by subtracting the mean.
• Start by testing the correlation between body temperature and weight using Pearson’s correlation
coefficient
• Build a model including both weight and sex, and give the code you used
• Report the goodness-of-fit (F-statistic and associated p-value) and the R2
• The model should provide you with three coefficients. Give a very brief interpretation of what
each of these mean (see the lecture notes for an example)
• Build a subset model using only weight and then use the anova() function to test whether or
not there is a significant improvement in the full model over the subset. Give the F-statistic, the
associated p-value and state whether or not you believe the full model to be better
7
setwd ("C:/doug/classes/geog6000")
temp<- read.csv("normtemp.csv", row.names=1, na.strings="")
temp$sex <- factor(temp$sex, labels=c("male","female"))what does the data frame look like?
summary(temp)## temp sex weight
## Min. : 96.30 male :65 Min. :57.00
## 1st Qu.: 97.80 female:65 1st Qu.:69.00
## Median : 98.30 Median :74.00
## Mean : 98.25 Mean :73.76
## 3rd Qu.: 98.70 3rd Qu.:79.00
## Max. :100.80 Max. :89.00temp$wtc <- temp$weight - (mean(temp$weight))did that work?
summary(temp$wtc)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -16.7615 -4.7615 0.2385 0.0000 5.2385 15.2385cor(temp$wtc, temp$temp)## [1] 0.2536564tempmodel <- lm(temp$temp ~ temp$wtc + temp$sex)
summary(tempmodel)##
## Call:
## lm(formula = temp$temp ~ temp$wtc + temp$sex)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.86363 -0.45624 0.01841 0.47366 2.33424
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 98.114528 0.087102 1126.428 < 2e-16 ***
## temp$wtc 0.025267 0.008762 2.884 0.00462 **
## temp$sexfemale 0.269406 0.123277 2.185 0.03070 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7017 on 127 degrees of freedom
## Multiple R-squared: 0.09825, Adjusted R-squared: 0.08405
## F-statistic: 6.919 on 2 and 127 DF, p-value: 0.001406the reduced subset model
attach(temp)## The following object is masked _by_ .GlobalEnv:
##
## tempsubsetmodel <- lm(temp ~ wtc, data = temp)
summary (subsetmodel)##
## Call:
## lm(formula = temp ~ wtc, data = temp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.85017 -0.39999 0.01033 0.43915 2.46549
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 98.249231 0.062444 1573.402 < 2e-16 ***
## wtc 0.026335 0.008876 2.967 0.00359 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.712 on 128 degrees of freedom
## Multiple R-squared: 0.06434, Adjusted R-squared: 0.05703
## F-statistic: 8.802 on 1 and 128 DF, p-value: 0.003591anova(subsetmodel)## Analysis of Variance Table
##
## Response: temp
## Df Sum Sq Mean Sq F value Pr(>F)
## wtc 1 4.462 4.4618 8.8021 0.003591 **
## Residuals 128 64.883 0.5069
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1