statedata <- read.csv("statedata.csv")
statedata2 <- statedata[,-1]mod.0 <-lm(Life.Exp ~ 1, data = statedata2)mod.1<-lm(Life.Exp ~., data =statedata2)step(mod.0, scope = formula(mod.1))## 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 15.528##
## Call:
## lm(formula = Life.Exp ~ Murder + HS.Grad + Frost + Population,
## data = statedata2)
##
## Coefficients:
## (Intercept) Murder HS.Grad Frost Population
## 7.103e+01 -3.001e-01 4.658e-02 -5.943e-03 5.014e-05The stepwise function steps forward through the variables and reduces the AIC at each step. The AIC is -28.16 compared to the original AIC of 30.44. The final model consists of Murder, HS.GRad, Frost and Population. These results indicate that life expectancy increases as high school graduates and population increases and decreases when there’s a change in murder and frost.
mod<-lm(formula = Life.Exp ~ Murder + HS.Grad + Frost + Population,
data = statedata2)
summary(mod)##
## Call:
## lm(formula = Life.Exp ~ Murder + HS.Grad + Frost + Population,
## data = statedata2)
##
## 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-12Using the summary() function, the goodness-of-fit reported an F-statistic of 31.37,p-value of 1.69e-12, and r2 value of 0.71256
newstate <- statedata[44,]
newstate[2] <- 2785
newstate[7] <- 75
newstate[6] <- 1.3
predict(mod, level = 0.95, interval = "conf", newstate)## fit lwr upr
## 44 73.45601 72.83971 74.07231Predicted life expectancy in Utah, with the above conditions is 73 years, with a 95% confidence level of 72.6 to 73.5.
newstateca <- statedata[5,]
newstateca[2] <- 36962
newstateca[7] <- 68.3
newstateca[6] <- 5.3
predict(mod, level = 0.95, interval = "conf", newstateca)## fit lwr upr
## 5 74.35232 72.64674 76.05789Predicted life expectancy in California, with the above conditions is 74 years, with a 95% confidence level of 71.5 to 74.6
bodytemp<- read.csv("normtemp.csv")
bodytemp$sex <- factor(bodytemp$sex, labels = c("male", "female"))
bodytemp$weight2 = bodytemp$weight - mean(bodytemp$weight)cor(bodytemp$temp,bodytemp$weight2)## [1] 0.2536564cor.test(bodytemp$temp,bodytemp$weight)##
## Pearson's product-moment correlation
##
## data: bodytemp$temp and bodytemp$weight
## t = 2.9668, df = 128, p-value = 0.003591
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.08519113 0.40802170
## sample estimates:
## cor
## 0.2536564Correlation is 0.25 and a p-value of 0.003591. From the Pearson’s correleation test we can reject the null hypothesis that the correlation is equal to 0.
bodytemp.lm3 <- lm(bodytemp$temp ~ bodytemp$weight2 + bodytemp$sex, data =bodytemp)
summary(bodytemp.lm3)##
## Call:
## lm(formula = bodytemp$temp ~ bodytemp$weight2 + bodytemp$sex,
## data = bodytemp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.86363 -0.45624 0.01841 0.47366 2.33424
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 98.114528 0.087102 1126.428 < 2e-16 ***
## bodytemp$weight2 0.025267 0.008762 2.884 0.00462 **
## bodytemp$sexfemale 0.269406 0.123277 2.185 0.03070 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7017 on 127 degrees of freedom
## Multiple R-squared: 0.09825, Adjusted R-squared: 0.08405
## F-statistic: 6.919 on 2 and 127 DF, p-value: 0.001406F-statistic of 6.919, R2 value of 0.09 and p-value of 0.001406 means this model describes 9% of the variation in the data set and that both variables are statistically significantly related to body temperature.
The three coeffecienets from the model are Intercept (98.114528), Weight2(0.025) and Sex Female (0.269). The sex coeffiencent means that if the person is a female, their body temperature is 0.269 higher than a male of the same weight. If the sex is female, the weight will be adjusted up by 0.269.
bodytemp.lm4 = lm(bodytemp$temp~weight2, data= bodytemp)
summary(bodytemp.lm4)##
## Call:
## lm(formula = bodytemp$temp ~ weight2, data = bodytemp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.85017 -0.39999 0.01033 0.43915 2.46549
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 98.249231 0.062444 1573.402 < 2e-16 ***
## weight2 0.026335 0.008876 2.967 0.00359 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.712 on 128 degrees of freedom
## Multiple R-squared: 0.06434, Adjusted R-squared: 0.05703
## F-statistic: 8.802 on 1 and 128 DF, p-value: 0.003591anova(bodytemp.lm3,bodytemp.lm4)## Analysis of Variance Table
##
## Model 1: bodytemp$temp ~ bodytemp$weight2 + bodytemp$sex
## Model 2: bodytemp$temp ~ weight2
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 127 62.532
## 2 128 64.883 -1 -2.3515 4.7758 0.0307 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The F-statistic is 4.7758, and the p-value is 0.0307. The subset model is not significantly better. Even though the p-value is still signifanct, the R squared value is lower. Adding the sex variable does not drastically change the model performance.