A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The

Question

2 years ago

9

A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The

results are given below. Construct a 95% confidence interval for the mean difference between the before and after scores. Is there evidence to suggest the logic course improves abstract reasoning? You may assume that the differences for the dependent samples are normally distributed . Before 74, 83, 75, 88, 84, 63, 93, 84, 91, 77 After 73, 77, 70, 77, 74, 67, 95, 83, 84, 75
Note : define d = before - after, then đ = 3.7 and s = 4.95
Please sketch the rejection region and show computation for the test statistic

Mathematics

1 answer:

iren [92.7K] · Answer 1 · 2022-09-13T19:27:42+03:00

Answer:

1. confidence interval = (0.163, 7.237)

1. confidence interval = (0.163, 7.237)2. t = 2.366

1. confidence interval = (0.163, 7.237)2. t = 2.3663. critical value of one tailed test = 1.833

Step-by-step explanation:

before after dt(before - after) d²

74 73 1 1

83 77 6 36

75 70 5 25

88 77 11 121

84 74 10 100

63 67 -4 16

93 95 -2 4

84 83 1 1

91 84 7 49

77 75 2 4

∑dt = 37

d* = 37/10

since sample space = 10

d* = 3.7

s.d from the question = 4.95

df = 10 - 1 = 9

critical value at 0.05 significance

t(0.025 at df of 9) = ±2.262

marginal error computation:

= (2.262) × (4.945/√10)

= 2.262 × 1.5637

= 3.5370

confidence interval CI = d* + marginal error

= 3.7 ±3.5730

= (0.163, 7.237)

The logic course give an improvement on abstract reasoning. The confidence interval shows that the result is significant.

H₀: Цd = 0

H₁: Цd > 0

∝ = 0.05

t = (3.7 - 0)/(4.945/√10)

t = 3.7/1.564

t = 2.366

for a right tailed test at 0.05 significance, and df of 9, the critical value is 1.833

please refer to the attachment to see the rejection region.