Question

A survey was given to a random sample of 400 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 168 respondents said they were in favor of the plan. Determine a 95% confidence interval for the proportion of people who favor the tax plan, rounding values to the nearest thousandth.

Answer 1

The sample proportion of individuals who favor the tax plan is 168/400 = 0.42.

The standard error of the sample proportion is sqrt[(0.42)(0.58)/400] = 0.032.

Using a 95% confidence level, the critical value is 1.96.

The margin of error is 1.96 * 0.032 = 0.063.

The 95% confidence interval is 0.42 ± 0.063, which is (0.357, 0.483).

Therefore, we can be 95% confident that the true proportion of people who favor the tax plan is between 0.357 and 0.483.