a)
yes, data appears to be linear
b)
X | Y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
75 | 644 | 36.00 | 2640.38 | 308.31 |
76 | 646 | 25.00 | 2438.84 | 246.92 |
77 | 658 | 16.00 | 1397.61 | 149.54 |
78 | 669 | 9.00 | 696.15 | 79.15 |
79 | 675 | 4.00 | 415.53 | 40.77 |
80 | 689 | 1.00 | 40.76 | 6.38 |
81 | 698 | 0.00 | 6.84 | 0.00 |
82 | 699 | 1.00 | 13.07 | 3.62 |
83 | 714 | 4.00 | 346.53 | 37.23 |
84 | 719 | 9.00 | 557.69 | 70.85 |
85 | 727 | 16.00 | 999.53 | 126.46 |
86 | 744 | 25.00 | 2363.46 | 243.08 |
87 | 758 | 36.00 | 3920.69 | 375.69 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 1053 | 9040 | 182 | 15837.1 | 1688.00 |
mean | 81.00 | 695.38 | SSxx | SSyy | SSxy |
sample size , n = 13
here, x̅ = Σx / n= 81.00 ,
ȳ = Σy/n = 695.38
SSxx = Σ(x-x̅)² = 182.0000
SSxy= Σ(x-x̅)(y-ȳ) = 1688.0
estimated slope , ß1 = SSxy/SSxx = 1688.0
/ 182.000 = 9.2747
intercept, ß0 = y̅-ß1* x̄ =
-55.8681
so, regression line is Ŷ = -55.868
+ 9.275 *x
R² = (Sxy)²/(Sx.Sy) = 0.9885
98.85% of variation is explained by this line
c)
α= 0.05
t critical value= t α/2 =
2.201 [excel function: =t.inv.2t(α/2,df) ]
estimated std error of slope = Se/√Sxx =
4.06024 /√ 182.00 =
0.301
margin of error ,E= t*std error = 2.201
* 0.301 = 0.662
estimated slope , ß^ = 9.2747
lower confidence limit = estimated slope - margin of error
= 9.2747 - 0.662
= 8.61
upper confidence limit=estimated slope + margin of error
= 9.2747 + 0.662
= 9.94
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