Question

Find the sample size needed to estimate the proportion of people who play video games, with 90% confidence. Assume a previous poll shows that about 16% of people play video games. The error should be less than 3 percent points.

Critical Values: z0.005 = 2.575, z0.01 = 2.325, z0.025 = 1.96, z0.05 = 1.645

When d.f.=6: t0.005 = 3.707, t0.01 = 3.143, t0.025 = 2.447, t0.05 = 1.943

When d.f.=9: t0.005 = 3.250, t0.01 = 2.821, t0.025 = 2.262, t0.05 = 1.833

When d.f.=44: t0.005 = 2.690, t0.01 = 2.412, t0.025 = 2.014, t0.05 = 1.679

When d.f.=100: t0.005 = 2.625, t0.01 = 2.364, t0.025 = 1.984, t0.05 = 1.660

Answer 1

Solution,

Given that,

= 0.16

1 - = 1 - 0.16 = 0.84

margin of error = E = 0.03

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.03)2 * 0.16 * 0.84

= 404.09

sample size = n = 405