In: Statistics and Probability
Question #11 – Regression Analysis
Use the data provided to:
Use a level of significance of 5% (α = .05).
Clearly show the null and alternate hypothesis.
Graphs are not required.
X |
Y |
3 |
14 |
7 |
26 |
6 |
23 |
4 |
17 |
7 |
28 |
5 |
20 |
8 |
29 |
2 |
11 |
x | y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
3 | 14 | 5.06 | 49.00 | 15.75 |
7 | 26 | 3.06 | 25.00 | 8.75 |
6 | 23 | 0.56 | 4.00 | 1.50 |
4 | 17 | 1.56 | 16.00 | 5.00 |
7 | 28 | 3.06 | 49.00 | 12.25 |
5 | 20 | 0.06 | 1.00 | 0.25 |
8 | 29 | 7.56 | 64.00 | 22.00 |
2 | 11 | 10.56 | 100.00 | 32.50 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 42 | 168 | 31.500 | 308.0 | 98.00 |
mean | 5.25 | 21.00 | SSxx | SSyy | SSxy |
sample size , n = 8
here, x̅ = Σx / n= 5.25 ,
ȳ = Σy/n = 21.00
SSxx = Σ(x-x̅)² = 31.5000
SSxy= Σ(x-x̅)(y-ȳ) = 98.0
estimated slope , ß1 = SSxy/SSxx = 98.0
/ 31.500 = 3.1111
intercept, ß0 = y̅-ß1* x̄ =
4.6667
so, regression line is Ŷ =
4.6667 + 3.1111 *x
SSE= (SSxx * SSyy - SS²xy)/SSxx =
3.111
std error ,Se = √(SSE/(n-2)) =
0.720
........
slope hypothesis test
tail= 2
Ho: ß1= 0
H1: ß1╪ 0
n= 8
alpha = 0.05
estimated std error of slope =Se(ß1) = Se/√Sxx =
0.720 /√ 31.50 =
0.1283
t stat = estimated slope/std error =ß1 /Se(ß1) =
3.1111 / 0.1283 =
24.2487
t-critical value= 2.447 [excel function:
=T.INV.2T(α,df) ]
Degree of freedom ,df = n-2= 6
decison : test stat > critical value , reject Ho
Conclusion: Reject Ho and conclude that slope is
significantly different from zero
so regression is significant
...............
Predicted Y at X= 4
is
Ŷ = 4.66667 +
3.111111 * 4 =
17.111