With a perfectly balanced roulette wheel, in the long run, red numbers should turn up 18 times in 38. To test its wheel, one casino records the result of 3,800 plays, finding 1,890 red numbers. Is that too many reds? Or chance variation?

(a) Formulate the null and alternative hypothesis as statements about a box model.

(b) The null says that the percentage of reds in the box is _. The alternative says that the percentage of reds in the box is __. Fill in the blanks.

(c) Compute z and P.

(d) Were there too many reds?

1. We are counting the number of reds, so we have a box with "1" s for reds and "0"s for others. The question is what proportion of 1's and 0' are in the box.

a) The null hypothesis is that the wheel is balanced, so the box contains 18 "1"s and 20 "0"s. Or you might state this in terms of the proportion of 1's (18/38) and 0's (20/38). The alternative hypothesis if that the wheel favors reds, so the proportion of 1's is more than 18/38.

b) Under the null hyp., we expect 6000*(18/38)= 2842.105 Reds. The SE is sqrt(6000)* SD(Box)= Sqrt(6000)*(1-0)*sqrt((18/38)*(20/38))= 38.67615. We observed2806, which is fewer than we expected. The question is , is it much fewer that we suspect foul play, or it is the usual sort of variation we expect due to chance? The test statistic is z=(obs-expected)/SE =(2806-2842.105)/38.67615= -0.933521.

c) The p-value is the probability of gettinga test statistic as extreme or morer extreme than this, which in this case means seeing a test statistic less than or equal to - 0.933521. The closest value gives a probability of (100-65.79)/2 = 17.105%

d) If we reject H0 and claim the roullette wheel is unfair, then there's a 17.105% chance we are wrong. This is a large probability of making a serious mistake, and it would be reasonable to not reject and conclude that there is insufficient evidence to suspect the wheel is unfair. (we would reject H10 if the p-value were less than 5%. Since it is 17%, we won't reject)