Final answer:
The decision rule for a two-tailed hypothesis test with a significance level of 0.05 and a small sample size is to reject the null hypothesis if the Z-score is greater than 1.96 or less than -1.96.
Step-by-step explanation:
The question pertains to setting up a decision rule for a hypothesis test where the null hypothesis H0: π = 0.50 and the alternative hypothesis H1: π ≠ 0.50. Since the alternative hypothesis suggests a two-tailed test, you must consider both tails of the distribution for your decision rule. The significance level given is α = 0.05, meaning that you would reject the null hypothesis if the test statistic falls into the upper 2.5% or the lower 2.5% of the distribution, as 5% significance level split between two tails of the distribution gives 2.5% for each tail.
In a situation with a sample size of 9, a Z-test for proportions might be used if the normal approximation is appropriate. If the normal approximation isn't valid due to the small sample size, an exact test like Fisher's exact test might be more appropriate. Since the provided information does not indicate which test to use, we will assume a Z-test for the explanation. With an assumed Z-test, the decision rule is to reject H0 if the test statistic is greater than the Z-score corresponding to the upper 2.5% of the distribution (Z > Z₀ˣ₅) or if it's less than the Z-score corresponding to the lower 2.5% of the distribution (Z < -Z₀ˣ₅).
Using the standard normal distribution, the Z-scores that correspond to the upper and lower 2.5% are approximately 1.96 and -1.96, respectively. Therefore, if the calculated Z-score from your sample proportion is greater than 1.96 or less than -1.96, you would reject the null hypothesis at the 5% significance level.