Question2. (Section 6.3) [Testing Binomial Data]
Ｙou may want to begin this problem by writing down the definition of $\alpha]$. Suppose the null-hypothesis $H_{0}: p=0.7$ is tested against the alternative-hypothesis $H_{1}: p<0.7$ using a small sample size of $n=7$. If the decision rule is to "Reject $H_{0}$ if $k \leq 3$ ", then what is the test's level of significance $\alpha$ ?

Studdy Solution
Calculate the probability of observing $k \leq 3$ under the null hypothesis $H_0: p = 0.7$.
The random variable $X$ follows a binomial distribution $X \sim \text{Binomial}(n=7, p=0.7)$.
Calculate:
$$\alpha = P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$$
Use the binomial probability formula:
$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Calculate each probability:
$$P(X = 0) = \binom{7}{0} (0.7)^0 (0.3)^7 = 1 \times 1 \times 0.0002187 = 0.0002187$$
$$P(X = 1) = \binom{7}{1} (0.7)^1 (0.3)^6 = 7 \times 0.7 \times 0.000729 = 0.0035745$$
$$P(X = 2) = \binom{7}{2} (0.7)^2 (0.3)^5 = 21 \times 0.49 \times 0.00243 = 0.0245027$$
$$P(X = 3) = \binom{7}{3} (0.7)^3 (0.3)^4 = 35 \times 0.343 \times 0.0081 = 0.0972405$$
Sum these probabilities:
$$\alpha = 0.0002187 + 0.0035745 + 0.0245027 + 0.0972405 = 0.1255364$$
The test's level of significance $\alpha$ is:
$$\boxed{0.1255}$$