Question

EX1	EX2	Ex3	FINAL
73	80	75	152
93	88	93	185
89	91	90	180
96	98	100	196
73	66	70	142
53	46	55	101
69	74	77	149
47	56	60	115
87	79	90	175
79	70	88	164
69	70	73	141
70	65	74	141
93	95	91	184
79	80	73	152
70	73	78	148
93	89	96	192
78	75	68	147
81	90	93	183
88	92	86	177
78	83	77	159
82	86	90	177
86	82	89	175
78	83	85	175
76	83	71	149
96	93	95	192

The following data provides 3 ex scores and 1 final ex score. Using the data you are to create a multiple linear regression line to predict final ex scores.

#### a
What is the correct model for all three tests?

#### b
How accurate is the model and interpret the R^squared value.

#### c
Check the conditions for a linear regression.

#### d
Interpret the intercept and discuss its implications.

#### e
Interpret the EX3 esimate in context.

#### f
Right a results sentence reporting your findings

Answer 1

The given data contains three ex score and final ex

score. We can fit the multiple regression model by considering final score as a independent variable and ex 1,2 and 3 scores as independent variables.

Now we can fit regression model to predict the final ex score. For this first we import scores data in R.

> #import the data in R

> score=read.csv(file="C:/Users/shree/Desktop/score.csv",header = T)

> #fit the model

> model=lm(score$final~., data=score)

> summary(model)

Call:

lm(formula = score$ï..final ~ ., data = score)

Residuals:

    Min      1Q  Median      3Q     Max 

-3.7452 -1.6328 -0.2984  0.8046  7.3111 

Coefficients:

            Estimate    Std. Error t value   Pr(>|t|)    

(Intercept)  -4.3361     3.7642    -1.152    0.26230    

ex1           0.3559     0.1214     2.932    0.00796  ** 

ex2           0.5425     0.1008     5.379    2.46e-05 ***

ex3           1.1674     0.1030     11.333   2.08e-10 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.614 on 21 degrees of freedom

Multiple R-squared:  0.9897,          Adjusted R-squared:  0.9882 

F-statistic: 670.1 on 3 and 21 DF,     p-value: < 2.2e-16

In the above table we can see that at 5% level of significance all the three variables are significant.

Also we get adjusted R squre value 98%. Which indicate that the above model fitted very well.

a)

The correct model for all three tests is

Y = b0 + b1 x1 + b0 + b1 x2+b3x3.

That is,

final_score= -4.33+0.355*ex2+0.54*ex2+1.16*ex3

b)

The p vlue is less than 0.05 and the value of R- suare and adjusted R-squre is 0.98.Which indicate that 98% variation in response variable (final_score) explained by independent variables(ex1,ex2 and ex3).

c)

Now the conditions for a linear regression.

• The relation between final_score and score in ex 1,2 and 3 are linear approximately. We can use scatter plot to check this condition.

• Error are uncorrelated. and constant variance.

• No multicollinearity present in data.(Means independent variable are uncorrelated).

d)

The value of intercept in the model is -4.33 and its negative. This means that the expected value of final scores will be zero when all independent variables (ex1, ex2 and e3) are set to be zero.

R-Code

#import the data in R
score=read.csv(file="C:/Users/shree/Desktop/score.csv",header = T)
str(score)  #check structure of data
head(score)
summary(score)
#fit the model
model=lm(score$ï..final~ex1+ex2,data=score)
summary(model)