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
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
1 answer:
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
= (<u>0.163, 7.237)</u>
<u>The logic course give an improvement on abstract reasoning. The confidence interval shows that the result is significant.</u>
H₀: Цd = 0
H₁: Цd > 0
∝ = 0.05
t = (3.7 - 0)/(4.945/√10)
t = 3.7/1.564
<u>t = 2.366</u>
for a right tailed test at 0.05 significance, and df of 9, the critical value is <u>1.833</u>
please refer to the attachment to see the rejection region.
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