Question

Suppose that you’re watching the World Series. At the beginning of the Series, you believe that the Cubs are a better team than the Indians…. (Sorry, I must pause here. As an Indians’ fan, it hurts to write that.) Anyways, you believe that the probability that the Cubs will beat the Indians in a single game is at least 0.50. You decide to reject this hypothesis if the Cubs fail to win more than 2 games in the World Series. (For those unfamiliar with the World Series, the first team to win 4 games wins the Series.) Let’s assume that the probability of victory is the same for each game and that all games are independent.

a) Let your test statistic be X, the number of games that the Cubs win. What distribution does X have under the null hypothesis? (Note that you may not be familiar with the name of this distribution. Just give the pmf then.)

b) What is your probability of Type I error?

c) If the true probability of winning for each game for the Cubs is 0.25, what is your probability of Type II error?

Answer 1

Total number of matches n = 7

Let probability of winning for cubs be p.

a)

X ~ the number of games that the Cubs win

Then as all the 7 matches are independent then X follows binomial with n = 7 and probability p as cubs have equal probability p to win each match. So there are 7 Bernoulli trials each with a winning probability of p and and this forms a binomial distribution.

X ~ B (7,p)

Probability Distribution Function is $\small P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}$

H₀ : $\small p \geq 0.5$

H₁ : $\small p < 0.5$

Critical Region $\small C = \left \{X \leq 2 \right \}$

b) By definition,

$\small \alpha = \underset{p \in H_0}{Max} \: \: \: p ( C\: | \: H_0 \: \: is \: \: true\: )$

$\small \alpha = P(X \leq 2 \: | \: p= 0.50)$

X has mean np and variance np(1-p).

Mean = 7*0.50 = 3.5

Variance = 7*0.5*0.5 = 1.75

By using Central Limit theorem,

$\small \alpha = P(\frac{X-\mu}{\sigma / \sqrt{n}} \leq \frac{2-\mu}{\sigma / \sqrt{n}} \: | \: p= 0.50)$

where $\small \mu = 3.5$

$\small \sigma = \sqrt{1.75}$

$\small n = 7$

$\small \alpha = P(Z \leq \frac{2-3.5}{\sqrt{1.75/7}} \:)$

$\small \alpha = P(Z \leq -3 \:)$

$\small \alpha = 0.00134$

c)

probability of type 2 error $\small \beta$ is given by:

$\small 1- \beta= \: p ( C\: | \: H_1 \: \: is \: \: true\: )$

given p =0.25

Mean = 7 * 0.25 = 1.75

Variance = 7 * 0.25 * 0.75 = 1.3125

By using Central Limit theorem,

$\small 1- \beta = P(\frac{X-\mu}{\sigma / \sqrt{n}} \leq \frac{2-\mu}{\sigma / \sqrt{n}} \: | \: p= 0.25)$

where $\small \mu = 1.75$

$\small \sigma = \sqrt{1.3125}$

$\small n = 7$

$\small 1-\beta = P(Z \leq \frac{2-1.75}{\sqrt{1.3125/7}} \:)$

$\small 1-\beta = P(Z \leq0.57735\:)$

$\small 1-\beta = 0.71814$

$\small \beta = 0.28186$ ( Probability Of Type - 11 error)