LearningLog4

Today in class we learned how to produce confidence and prediction intervals for a set of data. These are two important concepts because they help to identify how accurate a future estimate may be, or how sure we can be that a point lies close to the mean.

I begin my review by bringing in the BeerWings data I worked with previously.

library(resampledata)

## 
## Attaching package: 'resampledata'

## The following object is masked from 'package:datasets':
## 
##     Titanic

data("Beerwings")
attach(Beerwings)

beer.mod<- lm(Beer ~ Hotwings)
beer.mod

## 
## Call:
## lm(formula = Beer ~ Hotwings)
## 
## Coefficients:
## (Intercept)     Hotwings  
##       3.040        1.941

summary(beer.mod)

## 
## Call:
## lm(formula = Beer ~ Hotwings)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -18.566  -4.537  -0.122   3.671  17.789 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   3.0404     3.7235   0.817    0.421    
## Hotwings      1.9408     0.2903   6.686 2.95e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.479 on 28 degrees of freedom
## Multiple R-squared:  0.6148, Adjusted R-squared:  0.6011 
## F-statistic:  44.7 on 1 and 28 DF,  p-value: 2.953e-07

Above we see the summary of our data.

confint(beer.mod)

##                 2.5 %    97.5 %
## (Intercept) -4.586851 10.667590
## Hotwings     1.346131  2.535371

cor(Beerwings$Beer, Beerwings$Hotwings)

## [1] 0.7841224

We see the confidence interval above for our prediction of intercept and slope. We can also look at the information from our summary. A few key points to look at is our R-squared value of .6148, which is a significant measure that we can say our model explains 61% of the variation from our mean. With a smaller p-value, we can also expect less occurances of extremes measures from our mean. The last measure we had was a correlation component- which we can say that the two variables have a strong correlation with one another.

Now we will predict how many Hotwings were eaten if someone had dranken 30 FL OZ of beer.

myinfo<-data.frame(Hotwings=10)
predict(beer.mod, myinfo, interval= "predict")

##        fit      lwr      upr
## 1 22.44788 6.831517 38.06425

predict(beer.mod, myinfo, interval="confidence")

##        fit      lwr      upr
## 1 22.44788 19.42369 25.47207

Following the info received, if someone had eaten 10 wings, we are 95% sure the true mean lies within the interval 6.8-38.1 ounces of beer.

Extending this, we are 95% confident that the interval of 19.4-25.5 contains the true mean value of ounces drank by everyone who ate 10 wings.

Tankya for reading.