Question

Calculate the standard error. May normality be assumed? (Round your answers to 4 decimal places.)

		Standard Error		Normality
(a)	n = 22, ππ = .45			(Click to select)  No  Yes
(b)	n = 45, ππ = .18			(Click to select)  No  Yes
(c)	n = 110, ππ = .16			(Click to select)  Yes  No
(d)	n = 547, ππ = .003

Just need help with A-D, the standard error

Answer 1

A) n = 22

= 0.45

np = 22 * 0.45 = 9.9

nq = 22 * 0.55 = 22.1

Since np > 5 and nq > 5 so the normality may be assumed.

SE = sqrt(n * * (1 - ))

= sqrt(22 * 0.45 * 0.55) = 2.3334

B) n = 45

= 0.18

n = 45 * 0.18 = 8.1

n(1 - ) = 45 * 0.82 = 36.9

Since n > 5 and n(1 - ) > 5 so the normality may be assumed.

SE = sqrt(n * * (1 - ))

= sqrt(45 * 0.18 * 0.82) = 2.5772

C) n = 110

   = 0.16

n = 110 * 0.16 = 17.6

n(1 - ) = 110 * (1 - 0.16) = 92.4

Since n > 5 and n(1 - ) > 5 so the normality may be assumed.

SE = sqrt(n * * (1 - ))

= sqrt(110 * 0.16 * 0.84) = 3.8450

D) n = 547

   = 0.003

n = 547 * 0.003 = 1.641

n(1 - ) = 547 * (1 - 0.003) = 545359

Since n < 5, so the normality may not be assumed.

SE = sqrt(n * * (1 - ))

= sqrt(547 * 0.003 * (1 - 0.003)) = 1.2791