Question

Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us.

(a) Suppose n = 30 and p = 0.38. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

---Select--- Yes No , p̂  ---Select--- can cannot be approximated by a normal random variable because  ---Select--- nq does not exceed nq exceeds both np and nq exceed np exceeds np and nq do not exceed np does not exceed  .

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μ_p̂ =

σ_p̂ =

(b) Suppose n = 25 and p = 0.15. Can we safely approximate p̂ by a normal distribution? Why or why not?
---Select--- Yes No , p̂  ---Select--- can cannot be approximated by a normal random variable because  ---Select--- np exceeds both np and nq exceed np does not exceed nq exceeds nq does not exceed np and nq do not exceed  .

(c) Suppose n = 43 and p = 0.22. Can we approximate p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

---Select--- Yes No , p̂  ---Select--- can cannot be approximated by a normal random variable because  ---Select--- both np and nq exceed nq exceeds np does not exceed np and nq do not exceed np exceeds nq does not exceed  .

What are the values of μ_p̂ and σ_p̂? (Use 3 decimal places.)

μ_p̂ =

σ_p̂ =

Answer 1

a) n =30 and p =0.38

np = 11.4
nq = 18.6

Yes, p̂ can be approx normal distribution nq does not exceed nq exceeds both np and nq exceed

μp̂ = p = 0.38
σp̂ = sqrt( pq /n ) = sqrt( 0.38*0.62/30) = 0.088

b)
n =25 and p =0.15

np = 3.75
nq = 21.25

No , p̂ annot be approximated by a normal random variable

np<5

c)
n =43 and p =0.22

np = 9.46
nq = 33.54

Yes, p̂ can be approx normal distribution nq does not exceed nq exceeds both np and nq exceed 5

μp̂ = p = 0.22
σp̂ = sqrt( pq /n ) = sqrt( 0.22*0.78/30) = 0.0756