Answer:
(a) Both populations are Normally distributed.
Step-by-step explanation:
Hello!
There are two types of shields in older tractors, bolt-on shield, and flip-up shield. The researcher wants to study the proportion of older tractors whose shields were removed. For these two random samples where taken, one of older tractors with bolt-on shields and the other one of older tractors with flip-up shields, the number of tractors whose shields were removed was recorded in both groups.
Sample 1: bolt-on
nb= 183
xb= 35 with removed shields
pb'= 35/183= 0.19
Sample 2: flip-up
nf= 136
xf= 15 with removed shields
pf'= 15/136= 0.11
Statistic hypotheses:
H₀: pb=pf
H₁: pb≠pf
(a) Both populations are Normally distributed. d) The counts of successes and failures are large enough to use Normal calculations.
The populations are not normal, these are two binomial populations.
To be able to use the normal calculations the following conditions should be met:
Population 1
n₁≥30 ⇒ nb= 183
n₁*p₁'≥5 ⇒ nb*pb'= 183*0.19= 34.77
n₁*(1-p₁')≥5 ⇒ nb*(1-pb')= 183*(1-0.19)= 148.23
Population 2
n₂≥30 ⇒ nf= 136
n₂*p₂'≥5 ⇒ nf*pf'= 136*0.11= 14.96
n₂*(1-p₂')≥5 ⇒ nf*(1-pf')= 136*(1-0.11)= 121.04
(b) The data come from two independent samples (c) Both samples were chosen at random.
Both conditions are met, the samples are randomly selected, the randomization included in the sampling method guarantees the independence between the samples.
(e) Both populations are at least 10 times the corresponding sample sizes
Without the information regarding the populations, you'll have to assume that the population of old tractors with bolt-on shields is greater than 1830 and the population of old tractors with flip-up shields is greater than in 1360.
I hope this helps!
