9. The manager of a Casino has installed a game machine where the true proportio
ID: 1164022 • Letter: 9
Question
9. 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? (8 points)
10. Refer to question 9. 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? (8 points)
ANSWER QUESTION 10
Explanation / Answer
10) p= proportion of games won by people= 25/400= 0.0625
z score for 98% confidence level is 2.33
Confidence interval---
p - z*[p(1-p)/n]^0.5 to p + z[p(1-p)]^0.5
0.0625 - 2.33[0.0625*0. 9375]^0.5 to 0.0625 + 2.33[0.0625*0.0375]^0.5
0.074 to 0.199
This is the confidence interval
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