To determine which sample result provides the strongest evidence against the null hypothesis \( H_0: \mu \geq 30 \) in favor of the alternative hypothesis \( H_1: \mu < 30 \), you would look for a sample mean \( \bar{x} \) that is significantly less than 30.
In hypothesis testing, larger samples and lower sample means (when considering \( H_1: \mu < 30 \)) provide stronger evidence against the null hypothesis.
Since you've mentioned a sample size of 36, you might consider the standard error of the mean (SEM). If you have the population standard deviation \( \sigma \), you can calculate the SEM using:
\[ SEM = \frac{\sigma}{\sqrt{n}} \]
Where \( n \) is the sample size. Compare the sample means against the hypothesized mean (30), and the further the sample mean is below 30, the stronger the evidence for rejecting \( H_0 \).
However, without specific numeric values for the sample results or the population standard deviation, I can't provide a definitive answer.
If you provide the possible sample mean results or values from a set of sample results, I can analyze those to determine which provides the strongest evidence against the null hypothesis.