OpenIntro Here is OpenIntro’s RLab for Chapter 9: http://htmlpreview.github.io/?https://github.com/andrewpbray/oiLabs-base-R/blob/master/multiple_regression/multiple_regression.html Record and turn in...

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OpenIntro


Here is OpenIntro’s RLab for Chapter 9:



http://htmlpreview.github.io/?https://github.com/andrewpbray/oiLabs-base-R/blob/master/multiple_regression/multiple_regression.html




Record and turn in your answers for the Exercises.




  • In questions that ask for a number, give the numerical answer along with what that number describes. Be specific.




  • In questions that ask for your code input, give the code and also describe what R returns.




  • In questions that ask for explanation or analysis, provide a reasonably detailed response with evidence supporting your claims.



Answered Same DayJun 11, 2021

Answer To: OpenIntro Here is OpenIntro’s RLab for Chapter 9:...

Biswajit answered on Jun 18 2021
141 Votes
Assignment answers :
Question 1.This is an observational study as no Design of experiment was conducted but data collected as an observational study.Yes,it is possible to answer the question at primary level through data analysis.But more confirmatory will be a post facto design of experiment.
Questio
n 2:
This is a left skewed or negatively skewed distribution which shows that many students give very low score also.
Min. 1st Qu. Median Mean 3rd Qu. Max.
2.300 3.800 4.300 4.175 4.600 5.000
Here mean < median < mode.
Ideally we should see a normal distribution which is symmetrical in shape with central tendency parameters such as mean,median,mode around 2.5 to 3
Skewness of the distribution is -.699
Question 3 :
Question 4.
dim(evals)
[1] 463 21
There are 463 data points but seems fewer points on the scatter plot on the scatter plot below.
    Jitter adds a small amount of random variation to the location of each point, and is a useful way of handling overplotting caused by discreteness in smaller datasets.Earlier seemed like less no of points with smooth plot & overlapping.
Jitter added
Question 5:
m_bty
Call:
lm(formula = evals$score ~ evals$bty_avg)
Coefficients:
(Intercept) evals$bty_avg
3.88034 0.06664
summary(m_bty)
Call:
lm(formula = evals$score ~ evals$bty_avg)
Residuals:
Min 1Q Median 3Q Max
-1.9246 -0.3690 0.1420 0.3977 0.9309
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.88034 0.07614 50.96< 2e-16 ***
evals$bty_avg 0.06664 0.01629 4.09 5.08e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.5348 on 461 degrees of freedom
Multiple R-squared: 0.03502,    Adjusted R-squared: 0.03293
F-statistic: 16.73 on 1 and 461 DF, p-value: 5.083e-05
Score = 3.88 + .06 x bty_avg
The regression statistically significant as p value <.05 but the trendline shows it is practically not significant as if you have points of score uniformly over all the values of avg_bty.
Question 6 :
Residuals seem to be randomly distributed which is good.our red line follows our fitted line.
Standard residuals are right skewed approximately but otherwise approximately normal.
Our data seems approximately homoscedastic which fulfills uniform variance across range assumption of regression.
    It shows few leverage points i.e outliers in bty_avg data points.
We do not seem to have any influential points as measured by cook’s distance.
Question 7 :
Relationship is strong & correlation coefficient is 0.844
Correlation matrix plot shows strong relationship between the bty_ variables.
Multiple linear regression :
m_bty_gen <- lm(score ~ bty_avg + gender, data = evals)
> summary(m_bty_gen)
Call:
lm(formula = score ~ bty_avg + gender, data = evals)
Residuals:
Min 1Q Median 3Q Max
-1.8305 -0.3625 0.1055 0.4213 0.9314
Coefficients:
Estimate Std. Error t value Pr(>|t|) ...
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