Question

The manager of a Casino has installed a game machine where the true proportion that the player will win is set at .10. The manager monitors the machine by taking a sample of 144 games played. She will adjust the machine if the sample winning proportion is either less than .08 or more than .13 (and will not adjust if it is between .08 and .13). What is the probability she will have to adjust if the population proportion is in fact .10?

Refer to question above. Suppose that the Casino manager installs another different machine, but she is unsure of the true proportion of games that players will win. In a sample of 400 games, she found that players won 25 games. Compute a 98% confidence interval for the population proportion?

Answer 1

p = = 0.10

= sqrt(p(1 - p)/n)

      = sqrt(0.1 * 0.9/144) = 0.025

a) P(machine will adjust) = 1 - P(machine will not adjust)

                                        = 1 - (P(0.08 < < 0.13))

                                        = 1 - (P((0.08 - )/ < ( - )/ < (0.13 - )/))

                                        = 1 - (P((0.08 - 0.10)/0.025 < Z < (0.13 - 0.10)/0.025))

                                        = 1 - (P(-0.8 < Z < 1.2))

                                        = 1 - (P(Z < 1.2) - P(Z < -0.8))

                                        = 1 - (0.8849 - 0.2119)

                                        = 1 - 0.6730

                                        = 0.3270

b) = 25/400 = 0.0625

At 98% confidence interval the critical value is z_0.01 = 2.33

The 98% confidence interval for population proportion is

+/- z_0.01 * sqrt((1 - )/n)

= 0.0625 +/- 2.33 * sqrt(0.0625 * (1 - 0.0625)/400)

= 0.0625 +/- 0.0282

= 0.0343, 0.0907