In: Statistics and Probability
Ad expenditures |
2 |
6 |
13 |
Sales |
4 |
17 |
25 |
5 |
19 |
27 |
|
6 |
21 |
29 |
|
a)
Ho: µ1=µ2=µ3
H1: not all means are equal
treatment | 2 | 6 | 13 | |||||
count, ni = | 3 | 3 | 3 | |||||
mean , x̅ i = | 5.000 | 19.00 | 27.00 | |||||
std. dev., si = | 1.0 | 2.0 | 2.0 | |||||
sample variances, si^2 = | 1.000 | 4.000 | 4.000 | |||||
total sum | 15 | 57 | 81 | 153 | (grand sum) | |||
grand mean , x̅̅ = | Σni*x̅i/Σni = | 17.00 | ||||||
( x̅ - x̅̅ )² | 144.000 | 4.000 | 100.000 | |||||
TOTAL | ||||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 432.000 | 12.000 | 300.000 | 744 | ||||
SS(within ) = SSW = Σ(n-1)s² = | 2.000 | 8.000 | 8.000 | 18.0000 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 9
df within = N-k = 6
mean square between groups , MSB = SSB/k-1 =
744/2= 372.0000
mean square within groups , MSW = SSW/N-k =
18/6= 3.0000
F-stat = MSB/MSW = 372/3= 124.00
Decision: p-value<α , reject null
hypothesis
there is enough evidence of significant mean difference among three
treatments
b)
eta square ,effect size = SSbet/SST= 0.9764 (large)
c)
x | y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
2 | 4 | 25.0000 | 169.0000 | 65.000 |
2 | 5 | 25.0000 | 144.0000 | 60.000 |
2 | 6 | 25.0000 | 121.0000 | 55.000 |
6 | 17 | 1.0000 | 0.0000 | 0.000 |
6 | 19 | 1.0000 | 4.0000 | -2.000 |
6 | 21 | 1.0000 | 16.0000 | -4.000 |
13 | 25 | 36.0000 | 64.0000 | 48.000 |
13 | 27 | 36.0000 | 100.0000 | 60.000 |
13 | 29 | 36.0000 | 144.0000 | 72.000 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 63.00 | 153.00 | 186.00 | 762.00 | 354.00 |
mean | 7.00 | 17.00 | SSxx | SSyy | SSxy |
Sample size, n = 9
here, x̅ = Σx / n= 7.000
ȳ = Σy/n = 17.000
SSxx = Σ(x-x̅)² = 186.0000
SSxy= Σ(x-x̅)(y-ȳ) = 354.0
estimated slope , ß1 = SSxy/SSxx = 354/186=
1.9032
intercept,ß0 = y̅-ß1* x̄ = 17- (1.9032 )*7=
3.6774
Regression line is, Ŷ= 3.677 +
( 1.903 )*x
d)
R² = (SSxy)²/(SSx.SSy) =
0.8842
Approximately 88.42% of variation in
observations of variable sales , is explained by variable Ad
expenditure