Question

Let p be the probability that a coin will fall heads in a single toss in order to test \(H_0: p = \frac{1}{2}\) against the alternate \(H_1: p = \frac{3}{4}\). The coin is tossed five times and \(H_0\) is rejected if 3 or more than 3 heads are obtained. Then, the probability of Type I error, is

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{16} \)
• C. \( \frac{81}{128} \)
• D. \( \frac{1}{4} \)

Hint:

For a binomial distribution with \(p=0.5\), the distribution is symmetric. Therefore, \(P(X \ge k) = P(X \le n-k)\). Here, \(P(X \ge 3) = P(X \le 2)\). Since the total probability is 1, and the distribution is symmetric, \(P(X \ge 3) + P(X \le 2) = 1\), which implies \(P(X \ge 3) = 1/2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A Type I error occurs when we reject the null hypothesis (\(H_0\)) when it is actually true. The probability of a Type I error is denoted by \(\alpha\), also known as the significance level or the size of the test.

Step 2: Key Formula or Approach:
\(\alpha = P(\text{Reject } H_0 | H_0 \text{ is true})\). The experiment follows a binomial distribution. Let X be the number of heads in 5 tosses. Then \(X \sim \text{Bin}(n, p)\). The rejection rule is \(X \ge 3\). We need to calculate \(P(X \ge 3)\) under the assumption that \(H_0\) is true, i.e., \(p = 1/2\).

Step 3: Detailed Explanation:
Under \(H_0\), the number of heads X follows a binomial distribution with \(n=5\) and \(p=1/2\).
The probability mass function (PMF) is \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).
\[ P(X=k) = \binom{5}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{5-k} = \binom{5}{k} \left(\frac{1}{2}\right)^5 = \frac{\binom{5}{k}}{32} \]
The probability of a Type I error is the probability of rejecting \(H_0\), which is \(P(X \ge 3)\).
\[ \alpha = P(X \ge 3) = P(X=3) + P(X=4) + P(X=5) \] Calculate each probability:
- \( P(X=3) = \frac{\binom{5}{3}}{32} = \frac{10}{32} \)
- \( P(X=4) = \frac{\binom{5}{4}}{32} = \frac{5}{32} \)
- \( P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32} \)
Sum the probabilities: \[ \alpha = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]
Step 4: Final Answer:
The probability of Type I error is \( \frac{1}{2} \).