Question

There are two coins – one unbiased with P(H) = 1/2, the other biased with P(H) = 2/5
One of these coins is selected and tossed 5 times. If the head comes up at least twice,
the coin is assumed to be unbiased. Find the level of significance and power of the
test.

Answer 1

A decision rule is a procedure that the researcher uses to decide whether to accept or reject the null hypothesis.

Two types of errors can result from a decision rule. 

Type I error. A Type I error occurs when the researcher rejects a null hypothesis when it is true. The probability of committing a Type I error is called the significance level, and is often denoted by α. 

Type II error. A Type II error occurs when the researcher accepts a null hypothesis that is false. The probability of committing a Type II error is called Beta, and is often denoted by β. The probability of not committing a Type II error is called the Power of the test.

We have Binomial distribution for every coin

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Null hypothesis is that coin is unbiased (head comes up at least twice). Alternative hypothesis is that coin is biased.

Level of significance is probability to reject true null hypothesis:

$$\alpha=P_I(less\ than\ two\ heads)=P_I(0\ heads)+P_I(1\ heads)$$

$$\alpha=\dbinom{5}{0}({1\over 2})^0(1-{1 \over 2})^{5-0}+\dbinom{5}{1}({1\over 2})^1(1-{1 \over 2})^{5-1}$$

$$\alpha={1 \over 32}+{5 \over 32}={3 \over 16}=0.1875$$

Power of the test is the probability to reject null hypothesis while alternative is true:

$$1-\beta=P(less\ than\ two\ heads)=P(0\ heads)+P(1\ heads)$$

$$1-\beta=\dbinom{5}{0}({2\over 5})^0(1-{2 \over 5})^{5-0}+\dbinom{5}{1}({2\over 5})^1(1-{2 \over 5})^{5-1}$$

$$1-\beta={243 \over 3125}+{810 \over 3125}={1053 \over 3125}=0.33696$$