Answer:
Test statistic = 0.63
Pvalue = 0.555
A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
Explanation:
Given :
Before 13 22 65 123 56 63
After_ 14 21 43 84 75 72
To perform a paired t test :
H0 : μd = 0
H1 : μd ≠ 0
We obtain the difference between the two dependent sample readings ;
Difference, d = -1, 1, 22, 39, -19, -9
The mean of difference, Xd = Σd/ n = 33/6 = 5.5
The standard deviation, Sd = 21.296 (calculator).
The test statistic :
T = Xd ÷ (Sd/√n) ; where n = 6
T = 5.5 ÷ (21.296/√6)
T = 5.5 ÷ 8.6940555
T = 0.6326
The Pvalue : Using a Pvalue calculator ;
df = n - 1 = 6 - 1 = 5
Pvalue(0.6326, 5) = 0.5548
Decision region :
Reject H0 ; If Pvalue < α; α = 0.05
Since 0.5548 > 0.05 ; we fail to reject the Null and conclude that the data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.