Answer:
C) 150 men and 100 women
D) 200 men and 2000 women
E) 1000 men and 1000 women
Step-by-step explanation:
Hello!
To compare the proportion of people that carry certain genetic trait in men and woman from a certain population two variables of study where determined:
X₁: Number of men that carry the genetic trait.
X₁~Bi(n₁;p₁)
X₂: Number of women that carry the genetic trait.
X₂~Bi(n₂;p₂)
The parameter of interest is the difference between the population proportion of men that carry the genetic trait and the population proportion of women that carry the genetic trait, symbolically: p₁-p₂
To be able to study the difference between the population proportions you have to apply the Central Limit Theorem to approximate the distribution of both sample proportions to normal.
<u><em>Reminder:</em></u>
Be a variable with binomial distribution X~Bi(n;p), if a sample of size n is taken from the population in the study. Then the distribution of the sample proportion tends to the normal distribution with mean p and variance (pq)/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
So for both populations in the study, the sample sizes should be
n₁ ≥ 30
n₂ ≥ 30
Also:
Both samples should be independent and include at least 10 successes and 10 failures.
Both populations should be at least 20 times bigger than the samples. (This last condition is to be assumed because without prior information about the populations is impossible to verify)
- If everything checks out then (p'₁-p'₂)≈N(p₁-p₂; p(1/n₁+1/n₂))
<u>The options are:</u>
A) 30 men and 30 women
n₁ ≥ 30 and n₂ ≥ 30
Both samples are big enough and independent.
Population 1
Successes: x₁= n₁*p₁= 30*0.08= 2.4
Failures: y₁= n₁*q₁= 30*0.92= 27.6
Population 2
Successes: x₂= n₂*p₂= 30*0.5= 15
Failures: y₂= n₂*q₂= 30*0.5= 15
The second condition is not met.
B) 125 men and 20 women
n₁ ≥ 30 but n₂ < 30
Both samples are independent but n₂ is not big enough for the approximation.
C) 150 men and 100 women
n₁ ≥ 30 and n₂ ≥ 30
Both samples are big enough and independent.
Population 1
Successes: x₁= n₁*p₁= 150*0.08= 12
Failures: y₁= n₁*q₁= 150*0.92= 138
Population 2
Successes: x₂= n₂*p₂= 100*0.5= 50
Failures: y₂= n₂*q₂= 100*0.5= 50
All conditions are met, an approximation to normal is valid.
D) 200 men and 2000 women
n₁ ≥ 30 and n₂ ≥ 30
Both samples are big enough and independent.
Population 1
Successes: x₁= n₁*p₁= 200*0.08= 16
Failures: y₁= n₁*q₁= 200*0.92= 184
Population 2
Successes: x₂= n₂*p₂= 2000*0.5= 1000
Failures: y₂= n₂*q₂= 2000*0.5= 1000
All conditions are met, an approximation to normal is valid.
E) 1000 men and 1000 women
n₁ ≥ 30 and n₂ ≥ 30
Both samples are big enough and independent.
Population 1
Successes: x₁= n₁*p₁= 1000*0.08= 80
Failures: y₁= n₁*q₁= 1000*0.92= 920
Population 2
Successes: x₂= n₂*p₂= 1000*0.5= 500
Failures: y₂= n₂*q₂= 1000*0.5= 500
All conditions are met, an approximation to normal is valid.
I hope this helps!
a number (n) and 6 is (=) 8.
+ 6) = 8 Use the Distributive Property