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 = 39 and p = 0.27. 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 does not exceed np and nq do not exceed np exceeds np does not exceed nq exceeds  .

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 and nq do not exceed np does not exceed nq does not exceed np exceeds both np and nq exceed nq exceeds  .

(c) Suppose n = 46 and p = 0.16. 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--- np does not exceed nq exceeds np and nq do not exceed both np and nq 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)

np = 10.53

nq=28.47

Yes p̂   can be approximated by a normal random variable because   both np and nq exceed

μ_p̂ = 0.27

std error of proportion=σp=√(p*(1-p)/n)=

0.071

b)

np = 3.75

nq=21.25

No ,  p̂ cannot be approximated by a normal random variable because np does not exceed

c)

np = 7.36

nq=38.64

Yes p̂   can be approximated by a normal random variable because   both np and nq exceed

μ_p̂ =0.16

std error of proportion=σp=√(p*(1-p)/n)=

0.054