Question

The population proportion is 0.50. What is the probability that a sample proportion will be within +/- .01 of the population proportion for each of the following sample sizes? Round your answers to 4 decimal places. Use z-table.

a. n=100

b. n=200

c.n=500

d.n=1000

Answer 1

SOLUTION:

Given data,

= 0.50

(a) n = 100

P(-0.01 < - < 0.01)

= P [ (-0.01 / sqrt(0.50(1-0.50) / 100) < (- / sqrt((1-) / n) < (0.01 / sqrt(0.50(1-0.50) / 100)) ]

= P [-0.2 < Z < 0.2 ]

= P(Z < 0.2) - P(Z < -0.2)

= 0.57926-0.42074

= 0.1585

(b) n = 200

P(-0.01 < - < 0.01)

= P [ (-0.01 / sqrt(0.50(1-0.50) / 200) < (- / sqrt((1-) / n) < (0.01 / sqrt(0.50(1-0.50) / 200)) ]

= P [-0.28 < Z < 0.28 ]

= P(Z < 0.28) - P(Z < -0.28)

= 0.61026-0.38974

= 0.2205

(c) n = 500

P(-0.01 < - < 0.01)

= P [ (-0.01 / sqrt(0.50(1-0.50) / 500) < (- / sqrt((1-) / n) < (0.01 / sqrt(0.50(1-0.50) / 500)) ]

= P [-0.45 < Z < 0.45 ]

= P(Z < 0.45) - P(Z < -0.45)

= 0.67364-0.32636

= 0.3473

(d) n = 1000

P(-0.01 < - < 0.01)

= P [ (-0.01 / sqrt(0.50(1-0.50) / 1000) < (- / sqrt((1-) / n) < (0.01 / sqrt(0.50(1-0.50) / 1000)) ]

= P [-0.63 < Z < 0.63]

= P(Z < 0.63) - P(Z < -0.63)

= 0.73565-0.26435

= 0.4713

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