Answer:
We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80
Step-by-step explanation:
Ti-89 Calculator
Sample mean x = 31,5
Sample standard deviation s₂ = 8,35
Sample size n₂ = 11
Ti-84 Calculator
Sample mean y = 46,2
Sample standard deviation s₁ = 9,99
Sample size n₁ = 12
t(s) = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂
t(s) = 12,9 / √(99,8/12) + (69,72/11)
t(s) = 12,9 / √8,32 + 6,34
t(s) = 12,9 / 3,83
t(s) = 3,37
Test Hypothesis
Null Hypothesis H₀ y - x > 1,80
Altenative Hypothesis Hₐ y - x ≤ 1,80
We have a t(s) = 3,37
We need to compare with t(c) critical value for
t(c) α; n₁ +n₂-2 df = 12 +11 -2 df = 21
If we choose CI = 95 % then α = 5 % α = 0,05
From t-student table
t(c) = 1,72
t(s) = 3,37
t(s) >t(c)
t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %
