Question

can someone show the steps and explain c and d parts of the problem. I'm not really sure what these formulas are or how they work pertaining to the question. please explain this like im 5 years old so confused!

5. Consider a roulette wheel in a certain casino. (Recall that a Nevada roulette wheel nas up .. There are 18 red spaces, 18 black, and 2 green. In each spin of the wheel, the ball is equally likely to land in any of the 38 spaces, and the spins are independent of each other.) We are doubtful that the roulette wheel is fair; that is, we think the claim that the spaces are all equally likely is not true. We will test this by seeing if the ball lands in a space twice, too soon any of the 38 spaces). To perform the test, we will spin the wheel over and over until the ball lands in a space for the second time. We will reject the casino's claim that the wheel is fair, if this event takes three or fewer spins. (a) What is our null hypothesis? (1 point) Ho: Wheel s fawr (b) Under the null hypothesis, would the distribution of the number of spins until the ball lands in any of the spaces twice be negative binomial? Explain. (1 point) No. If we were specifyng one particular pocket, then yes, but we are not (c) What is the significance level of this test? spuns gwei Ho of lavere la long ini aspace twice in 3 or fewer (a points) = P(X=2) + P (X=3) where X=# 7 spins until 2 ball lands and hire in = (58)*+2x445) x 37 x38 a space : A Sunce could be any of 0.0776 W 7.76% . 10638 spaces. (d) What is the power of this test against the alternative hypothesis that the probability that the ball lands in one particular space is 7.5% and the probability that it lands in any of the other 37 spaces is 2.5%? (4 points) Power =P(X=2 or 3 I one space chance = 0.075 & rest are o 025) - [20.075)*+ 2% (0.0753x0,928)] + 37x60.025€ + 2X(0.025 (0.975)] = 0.08425 = 18-4257

Answer 1

(c) The significance level of the test in this case is the probability of landing the ball in the same space for the 2nd time, in 3 or fewer spins. This can happen in 2 spins or 3 spins.

So we can write $\alpha$ = P(X=2) + P(X=3), where X is the variable representing the number of spins until the ball lands in the same space for the second time.

Probability of selecting a single space out of the 38 equally likely spaces = 1/38

Probability of selecting a different space other than a designated single space = 1 - 1/38 = 37/38

Probability (X=2) = probability of selecting a single space x probability of the ball landing in the same slot x probability of the ball landing in the same slot = 38 x (1/38) x (1/38)=

38

P(X=3) = probability of selecting a single space x probability of the ball landing in the same slot x probability of the ball landing in the same slot x probability of the ball landing in a different x no. of possible combinations among the 3 spaces.

The last part,  no. of possible combinations among the 3 spaces can be (same space, different space, same space) or (different space, same space, same space). Hence this is equal to 2.

P(X=3) = 38 x (1/38) x (1/38) x (37/38) x 2

Hence, , or 7.76%

a = 38* )2* *2 = 0.0263 +0.0512 = 0.0776

(d) The power of the test is the above probability, given that the probability of the ball landing in a particular space is 0.075 and the probability of the ball landing in each of the other spaces is 0.025.

For the following calculations, the ball can either land in the biased space with probability 0.075 or it can land in one of the unbiased spaces with probability 0.025

Probability (X=2) = probability of the ball landing in the biased slot x probability of the ball landing in the biased slot + probability of selecting a single unbiased slot out of the 37 slots x probability of the ball landing in the same slot x probability of the ball landing in the same slot = 0.075 x 0.075 + 37 x 0.025 x 0.025 = 0.00563 + 0.02313 = 0.02875

P(X=3) = probability that the ball lands in the biased slot twice in 3 spins + probability that the ball lands in one of the unbiased slots in 3 spins = probability of ball landing in biased slot x probability of ball landing in biased slot x probability of ball landing in unbiased slot x no. of possible combinations (=2 as explained in part (c)) + probability of selecting an unbiased slot out of the 37 unbiased slots x probability of ball landing in that unbiased slot x probability of ball landing in the same unbiased slot x probability of ball landing in any of the other slots x no. of possible combinations (=2 as explained in part (c)) = 0.075 x 0.075 x (1-0.075) x 2 + 37 x 0.025 x 0.025 x (1-0.025) x 2 = 0.01041 + 0.04509 = 0.0555

Power = P(X=2) + P(X=3) = 0.02875 + 0.0555 = 0.08425, or 8.425%