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 = 43 and p = 0.13. 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 np does not exceed nq exceeds nq does not exceed np exceeds np and nq do 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--- cannot can be approximated by a normal random variable because  ---Select--- nq does not exceed np does not exceed both np and nq exceed np and nq do not exceed np exceeds nq exceeds  .

(c) Suppose n = 46 and p = 0.11. Can we approximate  p̂ by a normal distribution? Why? (Use 2 decimal places.)

np =

nq =

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

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

μ_p̂ =

σ_p̂ =

Answer 1

(a)

Since np = 43* 0.13 = 5.59 and n(1-p) = 37.41 both are greater than 5 so we can use normal approximation.

The sampling distribution of sample proportion will be approximately normal with mean

and standard deviation

(b)

Since np = 25* 0.15= 3.75 and n(1-p) = 21.5  both are not greater than 5 so we cannot use normal approximation.

The sampling distribution of sample proportion will be approximately normal with mean

and standard deviation

(c)

Since np = 46* 0.11= 5.06 and n(1-p) = 40.94 both are not greater than 5 so we can use normal approximation.

The sampling distribution of sample proportion will be approximately normal with mean

and standard deviation