Question

Suppose studies suggest that 52% of pets are overweight with 26% considered obese. Consider a vet who sees 49 random pets in the course of a day.

Step 1 of 3:

Describe the sampling distribution of the sample proportion of pets who are overweight if we use the pets that are seen in a single day as a random sample. Be sure to include all rationales.

Step 2 of 3:

Describe the sampling distribution of the sample proportion of pets who are overweight if we use the pets that are seen in a 5 day week as a random sample. Be sure to include all rationales.

Step 3 of 3:

Which of the following would accurately describe the sampling distribution of sample proportions for obese pets using a 5 day week as a random sample? Select the best answer.
μ_p = 52 and σ_p ≈ 0.028
μ_p = .26 and σ_p ≈ 0.028
CLT applies because our scenario is binomial such that n ≤ .05N and np(1-p) ≥ 10, therefore our sampling distribution is approximately normal.
σ_p ≈ 0.028
p = .26
μ_p = 52
μ_p = .26 and σ_p ≈ 0.028 and CLT applies because our scenario is binomial such that n ≤ .05N and np(1-p) ≥ 10, therefore our sampling distribution is approximately normal.
μ_p = .26 and σ_p ≈ 0.028 and CLT tells us it's not normal.
μ_p = .26

Answer 1

1) n = 49

    p = 0.52

= p = 0.52

= sqrt(p(1 - p)/n)

      = sqrt(0.52 * (1 - 0.52)/49)

      = 0.0714

2) n = 49 * 5 = 245

    p = 0.52

= p = 0.52

= sqrt(p(1 - p)/n)

      = sqrt(0.52 * (1 - 0.52)/245)

      = 0.0319

3) n = 245

    p = 0.26

= p = 0.26

= sqrt(p(1 - p)/n)

      = sqrt(0.26 * (1 - 0.26)/245)

     = 0.028

= 0.26 and = 0.028 and CLT applies because our scenario is binomial such that n < 0.05N and np(1 - p) > 10, therefore our sampling distribution is approximately normal.