Question

Question 4

A simple linear regression model was used in order to predict y, duration of relief from allergy, from x, dosage of medication. A total of n=10 subjects were given varying doses, and their recovery times noted. Here is the R output.

summary(lmod4)

##
## Call:
## lm(formula = y ~ x)
##
## Residuals:
##     Min      1Q Median      3Q     Max
## -3.6180 -1.9901 -0.4798 2.2048 3.7385
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept) -0.3997     2.7372 -0.146 0.887521   
## x             2.6399     0.4389   6.015 0.000318 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.807 on 8 degrees of freedom
## Multiple R-squared: 0.8189, Adjusted R-squared: 0.7963
## F-statistic: 36.18 on 1 and 8 DF, p-value: 0.000318

Compute 95% confidence intervals for β0 and β1. Give them in the form (Lower Bound, Upper Bound) NOT in the form estimate ± error margin. Also give the value of t that you used

β0: CI ( ________ , ________) ; β1: CI ( _________ , _________) ; t = __________

If the total sum of squares is TSS=348.05 and the residual sum of squares is RSS=63.02, find

R2= _____0.8189___________ ; Radj2 = _________0.7963___________

Answer 1

Coefficients:

Intercept β0 = -0.3997 with t value -0.146 and p valuee 0.887521. p value is relatively higher which means β0 is not significant enough. CI for β0 : (-3.1369 , 2.3375)

Coefficient β1 =2.6399 with t value 6.015 and p value 0.000318. this means this is highly significant.

CI = (2.201 , 3.079)

Model is: y = -0.3997 + 2.6399x which means with unit change in x, y changes 2.6399 times.

the Residual standard error: 2.807, meaning that the observed y values deviate from the predicted values by approximately2.807 units in average.

From the output above, the R2 is 0.8189, meaning that the observed and the predicted outcome values are highly correlated, which is very good for the data.

An Adjusted R2 that is close to 1 indicates that a large proportion of the variability in the outcome has been explained by the regression model whereas close to 0 0 indicates that the regression model did not explain much of the variability in the outcome. Adjusted R-squared: 0.7963 , which is good.

Larger F-statistic corresponds to a statistically significant p-value (if p < 0.05). Here the F-statistic: 36.18 on 1 and 8 DF, p-value: 0.000318, which is highly significant.

Model is fitted good.