In: Statistics and Probability
The data below are the final exam scores of 5 randomly selected calculus students and the number of hours they slept the night before the exam.
Hours, x | 4 | 6 | 3 | 9 | 3 |
Scores, y | 74 | 89 | 69 | 90 | 75 |
a) Draw scatterplot for the data.
b) Calculate the linear correlation coefficient to 3 decimal places.
(if you are unable to calculate the linear correlation coefficient, use .9 for part c,d and e)
c) Is there a linear relationship between the amount of sleep the student gets and the score on their exam. ( Justify your answer.Chart attached)
d) Determine the equation for the least squares regression line.
e) Predict the score of a student who studies 5 hours.
f) Interpret the slope and y-intercept of the least squares regression line
a)
b)
X | Y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
4 | 74 | 1 | 29.16 | 5.4 |
6 | 89 | 1 | 92.16 | 9.6 |
3 | 69 | 4 | 108.16 | 20.8 |
9 | 90 | 16 | 112.36 | 42.4 |
3 | 75 | 4 | 19.36 | 8.8 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 25 | 397 | 26 | 361.2 | 87 |
mean | 5 | 79.4 | SSxx | SSyy | SSxy |
correlation coefficient , r = Sxy/√(Sx.Sy)
= 0.8978
c)
sample size , n = 5
here, x̅ = 5 ȳ
= 79.4
SSxx = Σ(x-x̅)² = 26
SSxy= Σ(x-x̅)(y-ȳ) = 87
slope , ß1 = SSxy/SSxx =
3.34615
slope hypothesis test
tail= 2
Ho: ß1= 0
H1: ß1╪ 0
n= 5
alpha= 0.05
estimated std error of slope =Se(ß1) =
s/√Sxx =
0.9479
t stat = ß1 /Se(ß1) =
3.530058145
t-critical value=
3.182446305
p-value = 0.0386
decision : p-value<α , reject Ho
so, there is linear relation the amount of sleep the student gets
and the score on their exam
d)
slope , ß1 = SSxy/SSxx =
3.34615
intercept, ß0 = y̅-ß1* x̄ =
62.66923
so, regression line is Ŷ =
62.6692 + 3.3462 *x
e)
regression line is Ŷ =
62.6692 + 3.3462 *x
x=5
predicted score , Ŷ = 62.6692 + 3.3462 *5=79.4
f)
slope
for every unit increase in amount of sleeps in hours X, the score y will get increase by 3.35
intercept-when amount of sleeps is 0 hours, then score y is 62.67