Answer:
Calculated value t = 0.614< 1.782
we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .
Step-by-step explanation:
we will t-test t = x⁻- y⁻ /S√ n₁+n₂-2
Given An employer wishes to compare typing speeds of graduates from 2 different study programs: A and B
course A type at 62, 85, 59, 64, 73, 70, 75, and 72
mean of x is x⁻ = ∑x / n
mean of x = 62+85+59+64+73+70+75+ 72 /8 = 70
x⁻ = 70
course B type at 75, 64, 81, 55, 69, and 58
mean of y is y⁻ = ∑y / n = 75+64+81+55+69+ 58/6 = 67
y⁻ =67
course A course B
x y x- x⁻ (x-x⁻ )^2 y- y⁻ (y-y⁻ )^2
62 75 8 64 8 64
85 64 15 225 -3 9
59 81 -11 121 14 196
64 55 -6 36 -12 144
73 69 3 9 2 4
70 58 0 0 -9 81
75 5 25
72 2 4
∑( (x-x⁻ )^2= 484 ∑( (y-y⁻ )^2= 498
Adding ∑( (x-x⁻ )^2+ ∑( (y-y⁻ )^2= 982
n₁+n₂-2 = 8+6-2=12
now S^2 = 982/ 12 =81.833
Null hypothesis :H₀:μ₁=μ₂ ( there is no difference between A and B)
Alternative hypothesis :H₁:μ₁≠μ₂
level of significance ∝=0.05
we will use t-test t = 0.614 ( see attachment )
The degrees of freedom = n₁+n₂-2 = 8+6-2=12
From tabulated value t = 1.782
Calculated value t = 0.614< 1.782
we accepted null hypothesis for 12 degrees of freedom at 0.05 level of significance .
There is no difference between Course A and Course B
