Question

For a certain population of men, 8 percent carry a certain genetic trait. For a certain population of women, 0.5 percent carry the same genetic trait. Let pˆ1 represent the sample proportion of randomly selected men from the population who carry the trait, and let pˆ2 represent the sample proportion of women from the population who carry the trait. For which of the following sample sizes will the sampling distribution of pˆ1−pˆ2 be approximately normal?

Answer 1

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.

Reminder:

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))

The options are:

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!