Stat 141 R1 - Lecture #35
Announcements:
1) Assignment #11 Question 5:
The answer is wrong … should be “fail to reject” but MyStatLab wants
“reject”... so give the wrong answer for full marks in this question 
2) Exam:
STAT 141 R1 3 hrs 1400 Wed Apr 17 MAIN GYM,
~45 Multiple Choice Questions
Chapters 7, 8, 18-28 …. some pre MT skills will be required.
Simple Linear Regression …. continued
Last time:
Ex) Predicting final exam marks (%) from midterm exam marks (%) in a class
of 88 students:
Student
Midterm mark
Final mark
#1
67%
62%
#2
72%
50%
…
…
…
#88
88%
91%
Stat 141 R1 - Lecture #35
Given x = midterm percentage, y = final percentage,
𝑛 = 88, x̅ = 67.812, y̅ = 52.643,
sx = 17.922, sy = 25.430, r = 0.718
∑(yi–ŷi)2 = 27278.82
We had calculated:
The slope and intercept of the sample line of best fit:
 sample line of best fit: 𝑦̂= -16.443 + 1.019 x
An estimate for σ (standard deviation about the population line): se
Given SSE = ∑(yi–ŷi)2 = 27278.82 :
𝑠𝑒2 =
𝑆𝑆𝐸
𝑛−2
=
27278.82
88−2
= 317.196
→ 𝜎 ≈ 𝑠𝑒 = √317.196 = 17.810
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Stat 141 R1 - Lecture #35
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Inference for the population slope β1
When the 4 basic assumptions of the SLR model are satisfied:
o The relationship between x and y is sufficiently linear.
Presuming linearity, this means, at any x, με = 0.
o The std. dev. of ε is the same for any particular x (constant).
o The distribution of ε at any particular x is normal.
o The random deviations ε1, ε2, ..., εn associated with different
observations are independent of one another.
then: i) b1 is normally distributed
ii) The mean of b1 is 𝜇b1 = β1
iii) The standard deviation of b1 is 𝜎𝑏1 =
The standard error of b1 is SE(𝑏1 ) =
𝜎
√∑(𝑥𝑖 −𝑥̅ )2
𝑠𝑒
√∑(𝑥𝑖 −𝑥̅ )2
=
𝑠𝑒
𝑠𝑥 √𝑛−1
 CI for β1: b1± tα/2× SE(b1) with df= n– 2
Test statistic for H0: β1= 0 : 𝑡0 =
𝑏1 −𝛽1
𝑆𝐸(𝑏1 )
with df= n – 2
Ex) Construct a 95% CI for β1.
Sol.: SE(b1) =
𝑠𝑒
𝑠𝑥 √𝑛−1
=
17.810
17.922√87
= = 0.1065,
df = n-2 = 88-2 = 86 => using df = 75: tα/2 ≈ t0.025 = 1.992
b1± tα/2× SE(b1) = 1.019 ± (1.992)(0.1065) = 1.019 ± 0.212 =
(0.807, 1.231)
Stat 141 R1 - Lecture #35
page 4
Ex) Is there sufficient evidence to conclude that the final percentage
increases as midterm percentage increases? Carry out an appropriate
test using  = 0.01 .
Sol.: H0: β1=0
HA: β1> 0
Assumptions of SLR model: as above
Test statistic:
𝑏 −𝛽
1.019−0
𝑡0 = 1 1) =
= 9.568
𝑆𝐸(𝑏1
0.1065
with df = 88 – 2 = 86 => 75
P-value:
In the t-table, the corresponding range of p-values is (0.005, 0).
Note that the test is one-tailed.
Conclusion: Reject H0 in favour of HA at  = 0.01
There is convincing evidence against H0 in favour of HA: final
percentage increases as midterm percentage increases (a positive linear
association between mt% and fin%).
A typical summary table ( by Excel or StatsCrunch etc )
Stat 141 R1 - Lecture #35
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Inferences based on the estimated regression line:
• CI for the mean value of y corresponding to an x value:
For ŷν= b0 + b1xν
ŷν ± tdf, α/2× SE(𝜇̂ 𝜈 )
(df = n – 2)
• Prediction interval (PI) for an individual value of y corresponding to an
x value:
ŷν ± tdf, α/2× SE(𝑦̂𝜈 )
(df = n – 2)
Note that the PI is wider than the CI. Why?
Ex) Give a 95% CI for the mean final% when midterm% = 73%.
Compare with the 95% PI for final% when midterm% = 73%.
Sol.: ŷν= b0+ b1xν= -16.443+ 1.019(73) = 57.928%
where tn-2,α/2= t86,0.025 ≈ 1.992
Stat 141 R1 - Lecture #35
CI: ŷν ± tdf, α/2× SE(𝜇̂ 𝜈 )= 57.928 ± (1.992)(1.977) =
57.928 ± 3.939 = (53.990, 61.867)
PI: ŷν ± tdf, α/2× SE(𝑦̂𝜈 )= 57.928 ± (1.992)(17.919) =
57.928 ± 35.696 = (22.233, 93.624)
 THIS IS THE END! 
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