Question

City voters will soon go to the polls to decide whether to support a tax increase to build a new high school. Approval of bond issues like this requires a 60% "super majority" of yes votes. A local radio station phones 148 randomly selected voters, and finds 96 in favor of building the school. Create a 95% confidence interval.

Answer 1

The 95% confidence interval for the proportion of voters in favor of the tax increase to build a new high school is approximately 59.4% to 71.2%.

Step-by-step explanation:

To construct a 95% confidence interval for the proportion of voters in favor of the tax increase, we can use the formula for a confidence interval for a proportion:

`\[ \text{Confidence Interval} = \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]`

where

`\(\hat{p}\)` is the sample proportion (favorable votes / total votes),

(z) is the critical value corresponding to the desired confidence level (for 95%,
`\(z \approx 1.96\))`, and (n) is the sample size.

In this case, the sample proportion
`\(\hat{p}\)` is calculated as
`\((96)/(148) \approx 0.649\)` . The critical value (z) for a 95% confidence level is approximately 1.96, and the sample size (n) is 148.

Substituting these values into the formula, we get:

`\[ \text{Confidence Interval} = 0.649 \pm 1.96 \sqrt{(0.649(1-0.649))/(148)} \]`

Calculating this expression gives the confidence interval of approximately 59.4% to 71.2%.

This means that we can be 95% confident that the true proportion of voters in favor of the tax increase is within this interval. Therefore, there is evidence that the "super majority" requirement of 60% may be met, but it's not certain. The interval provides a range of plausible values for the true proportion based on the sample data.