A survey was given to a random sample of 215 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, $20 \%$ of the people said they were in favor of the plan. Determine a 95\% confidence interval for the percentage of people who favor the tax plan, rounding values to the nearest tenth.

Step 1 ：A survey was given to a random sample of 215 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 20% of the people said they were in favor of the plan.

Step 2 ：First, we calculate the sample proportion (p̂) which is the number of successes (people in favor of the plan) divided by the total number of trials (total people surveyed). In this case, 20% of 215 people were in favor of the plan.

Step 3 ：Next, we calculate the standard error of the proportion. The formula for the standard error of a proportion is \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), where n is the number of trials.

Step 4 ：Finally, we calculate the 95% confidence interval using the formula \(\hat{p} \pm Z \times SE\), where Z is the Z-score for a 95% confidence interval (1.96), \(\hat{p}\) is the sample proportion, and SE is the standard error.

Step 5 ：The 95% confidence interval for the percentage of people who favor the tax plan is \(\boxed{[0.1, 0.3]}\).