In hypothesis testing for a binomial distribution, Type I error ($\alpha$) occurs when the null hypothesis ($H_0$) is rejected even though it is true, and Type II error ($\beta$) occurs when the null hypothesis is accepted even though the alternative hypothesis ($H_1$) is true.

Here, the binomial distribution has:

• Number of trials ($n$) = 4,
• $H_0: p = \frac{1}{3}$,
• $H_1: p = \frac{1}{2}$,
• The test decision rule: Accept $H_0$ if $X \leq 2$, otherwise reject $H_0$.

Step 1: Type I Error ($\alpha$)

The probability of Type I error is calculated as: $\alpha = P(\text{Reject } H_0 | H_0 \text{ is true})$ Given that $H_0$ is true ($p = \frac{1}{3}$), we compute: $P(X > 2 | p = \frac{1}{3}) = 1 - P(X \leq 2 | p = \frac{1}{3})$

Using the binomial probability mass function: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

For $p = \frac{1}{3}$, compute $P(X = 0), P(X = 1), P(X = 2)$: $P(X = 0) = \binom{4}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^4 = 1 \cdot 1 \cdot \frac{16}{81} = \frac{16}{81}$ $P(X = 1) = \binom{4}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81}$ $P(X = 2) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81}$

Summing these probabilities: $P(X \leq 2 | p = \frac{1}{3}) = \frac{16}{81} + \frac{32}{81} + \frac{24}{81} = \frac{72}{81} = \frac{8}{9}$

Thus: $\alpha = 1 - \frac{8}{9} = \frac{1}{9}$

Step 2: Type II Error ($\beta$)

The probability of Type II error is: $\beta = P(\text{Accept } H_0 | H_1 \text{ is true})$ Given $H_1$ is true ($p = \frac{1}{2}$), compute: $P(X \leq 2 | p = \frac{1}{2})$

For $p = \frac{1}{2}$, compute $P(X = 0), P(X = 1), P(X = 2)$: $P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = 1 \cdot 1 \cdot \frac{1}{16} = \frac{1}{16}$ $P(X = 1) = \binom{4}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = 4 \cdot \frac{1}{2} \cdot \frac{1}{8} = \frac{4}{16}$ $P(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = 6 \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{6}{16}$

Summing these probabilities: $P(X \leq 2 | p = \frac{1}{2}) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} = \frac{11}{16}$

Thus: $\beta = \frac{11}{16}$

Final Results:

1. Type I error probability ($\alpha$): $\frac{1}{9}$
2. Type II error probability ($\beta$): $\frac{11}{16}$

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Would you like a detailed explanation of any step or related concept? Here are 5 follow-up questions you might consider:

1. How are binomial probabilities calculated using the PMF formula?
2. Can you explain the role of Type I and Type II errors in hypothesis testing?
3. What are some practical implications of minimizing $\alpha$ and $\beta$?
4. How does changing the decision rule affect $\alpha$ and $\beta$?
5. What is the relationship between $\alpha$, $\beta$, and the power of a test?

Tip: Always check that your decision rule aligns with the desired balance of risks for Type I and Type II errors.