Does the Public Support Sin Taxes? A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose' sin taxes' on soda and junk food" The proportion in faver of taxing these foods was 32 \% : \( { }^{1 * 32 \%} \) favor 'sin taves' on sods, Junk food", Rasmussen Reports, Aphll 1, 2011. Part 1 Your answer is correct. (a) Find a 90\% confidence interval for the proporition of US adults favoring taxes on soda and junk food. Round your answers to three decimal places. The \( 90 \% \) confidence interval is \( \square \) 296 to 344 eTextbook and Media Hint Attempts: unlimited Part 2 (b) What is the margin of error? Round your answer to one decimal place. margin of error = \( \square \) \%

To calculate the 90% confidence interval for the proportion of US adults favoring sin taxes on soda and junk food, we can begin by determining the sample proportion \(p\) and the sample size \(n\). The proportion of adults favoring the tax is \(p = 0.32\), and the sample size is \(n = 1000\). To find the standard error (SE), we can use the following formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.32(1-0.32)}{1000}} = \sqrt{\frac{0.32 \times 0.68}{1000}} = \sqrt{0.0002176} \approx 0.0147 \] For a 90% confidence interval, the critical z-value (z*) is approximately 1.645. The margin of error (ME) can be calculated as: \[ ME = z^* \times SE = 1.645 \times 0.0147 \approx 0.0242 \] Now we can calculate the confidence interval: \[ \text{Confidence Interval} = \hat{p} \pm ME = 0.32 \pm 0.0242 \] This gives us: \[ 0.32 - 0.0242 \approx 0.296 \quad \text{and} \quad 0.32 + 0.0242 \approx 0.344 \] So, the 90% confidence interval is \(0.296\) to \(0.344\). **Part 2 (b)**: The margin of error rounded to one decimal place is: \[ ME \approx 2.4\% \] Therefore, the final answers are: - The 90% confidence interval is **0.296 to 0.344** - The margin of error is **2.4%**.