Answer:
(8.734 ≤ μd ≤ 10.026)
Explanation:
Given the data:
Car Rear Front
1 41.6 32.6
2 35.8 26.7
3 46.4 37.9
4 46.2 36.9
5 38.8 29.9
6 51.8 42.3
7 51.2 42.5
8 44.1 33.9
9 47.3 36.1
Difference, d :
9, 9.1, 8.5, 9.3, 8.9, 9.5, 8.7, 10.2, 11.2
Mean difference, μd = Σd / n = 84.4 / 9 = 9.38
Standard deviation of difference, Sd = 0.84 (calculator)
The confidence interval :
μd ± margin of error
Margin of Error = Tcritical * Sd/√n
TCritical at 95%, df = 9-1 = 8
Tcritical = 2.306
Margin of Error = 2.306 * (0.84/√9) = 2.306*(0.84/3) = 0.64568
Lower boundary = 9.38 - 0.64568 = 8.73432
Upper boundary = 9.38 + 0.64568 = 10.02568
(8.734 ; 10.026)