Question

Can explain how to do these problems?

1 Type II Error: For the roulette table in (Q6), determine which hypothesis testing scenario has the larger Type II error probability for a two-sided hypothesis for HO: p=18/19:

1. a) N=10,000, p=0.96 , α=0.05 OR b) N=10,000, p=0.97 , α=0.05.

2. a) N=10,000, p=0.96, α=0.05 OR b) N=50,000, p=0.96, α=0.05.

3. a) N=10,000, p=0.97, α=0.05 OR b) N=10,000, p=0.97, α=0.01. Describe how the Type II error is influenced by N, p and α.

Q6:

Suppose that you are the manager of a casino and you collect data on one of your roulette tables. You find that among the last 10,000 bets, there were only 350 bets for which a red or black number did not win. On average, you expect the proportion of winning bets for black or red to equal p=18/19.

2. Hypothesis Testing: You collect data on N=100 individuals of similar height and weight who enter the Whole 30 diet program. You plan to conduct the following two-sided hypothesis test (α=0.01): H0: The average weight loss is 5 pounds (µ=5). HA: The average weight loss is not 5 pounds (µ≠5).

a) Suppose the true mean weight loss (µ) does equal 5 pounds. You compute a test statistic of Z = 2.2.

(i) State your decision: Reject H0 / Do Not Reject H0. (Circle your answer clearly)

(ii) Was an error made? Type I Error / Type II Error / Neither. (Circle your answer clearly)

b) Suppose the true mean weight loss (µ) equals 7 pounds. You compute a test statistic of Z = 2.2.

(i) State your decision: Reject H0 / Do Not Reject H0. (Circle your answer clearly)

(ii) Was an error made? Type I Error / Type II Error / Neither. (Circle your answer clearly)

Answer 1

6)

2)

a) for 0.01 level rejection region z<-2.576 or z>2.576

as test statistic falls in rejection region;

i)Reject Ho

ii)Type I error ; as we rejecting null when it is true)

b) std error =std deviation/(n)^1/2 =10/(100)^1/2 =1

for 0.05 level ; critcal vlaue of z =1.96

; acceptance region =sample mean -/+ z*std error = 5-/+ 1.96*1 =3.04 to 6.96

hence type II error =probability of not rejecting given mean is 6 =P(3.04<X<6.96)=P((3.04-6)/1<Z<(6.96-6)/1)

=P(-2.96<Z<0.96) =0.8315-0.0015 =0.8300

Please post the remaining questions in another post.