A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of 12 students using a Ti-84 calculator averaged 46.2 items identified, with a standard deviation of 9.99. Can we conclude that the mean number of times identified with the Ti-84 calculator exceeds that of the Ti-89 calculator by more than 1.80

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 %

a is the right answer

its 50 miles away from the starting point