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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 favor of taxing these foods was \( 32 \% \). 1" \( 32 \% \) favor 'sin taxes' on soda, junk food", Rasmussen Reports, April 1, 2011. Part 1 (a) Find a \( 90 \% \) confidence interval for the proportion of US adults favoring taxes on soda and junk food. Round your answers to three decimal places. The \( 90 \% \) confidence interval is i

Ask by Haynes Zimmerman. in the United States
Mar 18,2025

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The \(90\%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food is \(0.296\) to \(0.344\).

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**Step 1. Define the given values and the formula** We are given that the sample proportion is \[ \hat{p} = 0.32 \] and the sample size is \[ n = 1000. \] For a confidence interval for a proportion, we use the formula: \[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \] where \( z^* \) is the critical value corresponding to the desired confidence level. **Step 2. Determine the critical value** For a \(90\%\) confidence interval, the critical value is approximately \[ z^* = 1.645. \] **Step 3. Compute the standard error (SE)** The standard error is given by: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.32 \times 0.68}{1000}}. \] First, compute the product: \[ 0.32 \times 0.68 = 0.2176. \] Then, \[ SE = \sqrt{\frac{0.2176}{1000}} = \sqrt{0.0002176} \approx 0.01475. \] **Step 4. Compute the margin of error** The margin of error is: \[ ME = z^* \times SE = 1.645 \times 0.01475 \approx 0.02427. \] **Step 5. Construct the confidence interval** The confidence interval is: \[ \hat{p} \pm ME = 0.32 \pm 0.02427. \] Thus, the lower limit is: \[ 0.32 - 0.02427 \approx 0.29573, \] and the upper limit is: \[ 0.32 + 0.02427 \approx 0.34427. \] **Step 6. Round the answers** Rounding to three decimal places, the confidence interval is: \[ (0.296, \, 0.344). \]

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To calculate the \(90\%\) confidence interval for the proportion of US adults favoring sin taxes on soda and junk food, we can use the formula for the confidence interval for a proportion: \[ \text{CI} = \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Where: - \(\hat{p} = 0.32\) (the sample proportion in favor of the taxes), - \(n = 1000\) (the sample size), - \(z\) is the z-score corresponding to the confidence level (for \(90\%\), \(z \approx 1.645\)). Calculating the standard error (SE): \[ SE = \sqrt{\frac{0.32(1 - 0.32)}{1000}} = \sqrt{\frac{0.32 \times 0.68}{1000}} \approx \sqrt{0.0002176} \approx 0.0147 \] Now calculate the margin of error (ME): \[ ME = z \times SE = 1.645 \times 0.0147 \approx 0.0242 \] Now, construct the confidence interval: \[ \text{CI} = 0.32 \pm 0.0242 \] Calculating the lower and upper bounds: Lower bound: \[ 0.32 - 0.0242 \approx 0.2958 \text{ (rounded to 0.296)} \] Upper bound: \[ 0.32 + 0.0242 \approx 0.3442 \text{ (rounded to 0.344)} \] So, the \(90\%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food is approximately: \[ (0.296, 0.344) \]

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