Question

OTHER ANSWERS ONLINE ARE INCORRECT!! An American roulette wheel has 38 slots and 18 of these are red. If the wheel is fair, then the probability of it landing on red is 18/38. At a certain roulette table, after 190 spins it landed red on 78 of the spins. This is less than the expected value. Could chance explain the difference, or is this sufficient evidence otherwise? Use a hypothesis test as follows.

(a) Compute the expected value of the number of "reds" after 190 spins, assuming the null hypothesis (that the wheel is fair).

(b) Compute the SD of the number of "reds" after 190 spins, again assuming the null hypothesis. (You may round to two decimal places.)

(c) Compute the z-statistic. (Round to two decimal places.)

(d) What is the observed significance level? (Round to the nearest percentage. Use a Normal approximation; do not bother with a continuity correction.) %

(e) Is the result statistically significant, highly significant, or neither?

Answer 1

i.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 38 * 0.4736 = 17.9968
standard deviation ( √npq )= √38*0.4736*0.5264 = 3.0779
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
ii.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 190 * 0.4105 = 77.995
standard deviation ( √npq )= √190*0.4105*0.5895 = 6.7807
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
mean = 77.995
b.
standard deviation =6.7807
c.
Given that,
mean(x)=17.9968
standard deviation , sigma1 =3.0779
number(n1)=38
y(mean)=77.995
standard deviation, sigma2 =6.7807
number(n2)=190
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=17.9968-77.995/sqrt((9.47347/38)+(45.97789/190))
zo =-85.6
| zo | =85.6
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =85.599 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -85.6 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 != u2
test statistic: -85.60
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

d.
level of significance = 0.05
e.
we have enough evidence to support the claim that difference in means statistically significant