Respuesta :
Answer:
Step-by-step explanation:
Hello!
X₁: speed of a motorcycle at a certain intersection.
n₁= 135
X[bar]₁= 33.99 km/h
S₁= 4.02 km/h
X₂: speed of a car at a certain intersection.
n₂= 42 cars
X[bar]₂= 26.56 km/h
S₂= 2.45 km/h
Assuming
X₁~N(μ₁; σ₁²)
X₂~N(μ₂; σ₂²)
and σ₁² = σ₂²
A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.
The parameter of interest is μ₁-μ₂
(X[bar]₁-X[bar]₂)±[tex]t_{n_1+n_2-2}[/tex] * [tex]Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }[/tex]
[tex]t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654[/tex]
[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71[/tex]
[(33.99-26.56) ± 1.654 *([tex]3.71*\sqrt{\frac{1}{135} +\frac{1}{42} }[/tex])]
[6.345; 8.514]= [6.35; 8.51]km/h
Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15
[tex]t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485[/tex]
[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06[/tex]
[(475.12-321.34) ± 2.485 *([tex]33.06*\sqrt{\frac{1}{12} +\frac{1}{15} }[/tex])]
[121.96; 185.60]
I hope this helps!