In: Statistics and Probability
The number of hours 10 students spent studying for a test and their scores on that test is represented in the table below.
Hours spent studying, x |
0 |
2 |
4 |
5 |
5 |
5 |
6 |
7 |
8 |
9 |
Test scores, y |
40 |
51 |
64 |
69 |
73 |
75 |
93 |
90 |
95 |
95 |
7) What amount of the variance in test score (y) do the hours spent studying (x) account for this sample? (r2 =?) Interpret the result in the context of the data. (10p)
8) What is the standard error of the estimate (the unexplained variation around the regression line). (5p)
9) Would you conclude that hours of spent is a good predictor of a test score based upon this sample? What evidence do you have to support your conclusion? (10p)
x | y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
0 | 40 | 26.01 | 1190.25 | 175.95 |
2 | 51 | 9.61 | 552.25 | 72.85 |
4 | 64 | 1.21 | 110.25 | 11.55 |
5 | 69 | 0.01 | 30.25 | 0.55 |
5 | 73 | 0.01 | 2.25 | 0.15 |
5 | 75 | 0.01 | 0.25 | -0.05 |
6 | 93 | 0.81 | 342.25 | 16.65 |
7 | 90 | 3.61 | 240.25 | 29.45 |
8 | 95 | 8.41 | 420.25 | 59.45 |
9 | 95 | 15.21 | 420.25 | 79.95 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 51 | 745 | 64.9 | 3308.500 | 446.500 |
mean | 5.100 | 74.500 | SSxx | SSyy | SSxy |
7)
R² = (Sxy)²/(Sx.Sy) = 0.9285
92.85% amount of the variance in test score (y) do the hours spent studying (x) account for this sample
8)
SSE= (SSxx * SSyy - SS²xy)/SSxx =
236.663
std error ,Se = √(SSE/(n-2)) =
5.4390
9)
yes, we would conclude that hours of spent is a good predictor of a test score based upon this sample
because R² = 0.9285 is closer to 1