Answer:
Using Anova for a tri linear probability at ∝= 0.005
Step-by-step explanation:
Here simple probability cannot be used because we want to enter your answer as a tri-linear inequality using decimals (not percents).
So we can use ANOVA
Null hypothesis
H0: µA = µB=µC
all the means are equal
Alternative hypothesis
H1: Not all means are equal.
The significance level is set at α-0.005
The test statistic to use is
F = sb²/ sw²
Which if H0 is true has an F distribution with v₁=k-1 and v₂= n-k degrees of freedom .
The computations are as follows
XA (XA)² XB (XB)² XC (XC)² Total ∑X²
Male 19(361) 4(16) 12(144) 35 521
Female 3(9) 13 (169) 5 (25) 21 203
TotalTj 22 17 17 56 724
T²j (22)(22)
484 289 289 1062
∑X² 370 285 169
Correction Factor = CF = Tj²/n = (56)²/6= 522.67
Total SS ∑∑X²- C. F = 724- 522.67= 201.33
Between SS ∑T²j/r - C.F = 1062/ 2 - 522.877 =8.33
Within SS = Total SS - Between SS
=201.33- 8.33= 193
The Analysis of Variance Table is
Source Of Sum of Mean Computed
<u>Variation d.f Squares Squares F</u>
Between
<u>Samples 1 8.33 8.33 8.33/ 48.25= 0.1726</u>
Within
<u>Samples 4 193 48.25 </u>
The critical region is F >F ₀.₀₀₅ (1,4) = 31.3328
Calculated value of F = 0.1726
Since it is smaller than 5 % reject H0.
However the decimal probability will be
Male 19 4 12 35
Female 3 13 5 21
Total 22 17 17 56
There are total 22 people who get an A but only 19 males who get an A
So the probability that a male gets an A is = 19/22= 0.8636
