Question

FS2018 Home Announcements Syllabus Modules Assignments Grades People R Assignment 2 - Linear Regression Due Oct 26 at 11:59pm Time Limit None Points 59 Questions 2 Allowed Attempts Unlimited Instructions Use the R assignment and RStudio to answer the following questions project 2 script.R Note: what is below applies to p-values only! NOTE: P-values like 2e-7 is 0.0000002. If the p-value has a number beyond e-4, just write it as o. i.e. 3.2e-3-0.0032 1.04e-5-0 <2e-26 = 0 Write B1 0 as B/0 26

Answer 1

model=lm(data=x,temp~day+time+activ)
> summary(model)

Call:
lm(formula = temp ~ day + time + activ, data = x)

Residuals:
Min 1Q Median 3Q Max
-0.30893 -0.09317 0.01351 0.10232 0.26450

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -9.296e+01 1.971e+01 -4.716 7.11e-06
day 3.739e-01 5.684e-02 6.578 1.67e-09
time 2.849e-04 3.306e-05 8.620 5.55e-14
activ 2.533e-01 5.869e-02 4.316 3.49e-05

y=-9.296e+01 + 3.739e-01*day +2.849e-04 *time + 2.533e-01*activ

h0: b3=0 vs h1:b3 $\neq$ 0

alpha=0.01

now the value of t= 4.316

and t(0.005,110)=2.626 from t table

so 4.316> 2.626; we can reject the null hypothesis that b3=0, so activ is a significant predictor of temp.

> newdat=data.frame(day=346,time=1400,activ=0)
> predict(model,newdat)
1
36.7985

> newdat=data.frame(day=347,time=800,activ=1)
> predict(model,newdat, interval = 'confidence')
fit lwr upr
1 37.25471 37.13256 37.37686

> newdat=data.frame(day=347,time=1800,activ=1)
> predict(model,newdat, interval = 'prediction',level = 0.99)
fit lwr upr
1 37.53965 37.13619 37.94311

> confint(model,'time',level = 0.95)
2.5 % 97.5 %
time 0.0002194303 0.0003504472