Question

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 \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

Short answer

The estimated 95% confidence interval for the proportion of US adults favoring taxes on soda and junk food is approximately (0.29, 0.35). The margin of error is about 0.03. And roughly 9604 participants are needed in the survey for a margin of error of only 1% with 95% confidence.

Step-by-step solution

Calculate confidence interval

We first calculate the 95% confidence interval using the formula \(p \pm z*(\sqrt{(p(1-p))/n})\). Given, p = 0.32 (or 32%), z = 1.96 for a 95% confidence level, and n = 1000, we substitute these values into the formula and get the confidence interval.

Identify the margin of error

To find the margin of error, we just need to subtract the lower limit of our confidence interval in step 1 from the given proportion p = 0.32. This will give us the margin of error.

Calculate required sample size

The formula for margin of error is \(E = z * \sqrt{(p(1-p))/n}\). We can rearrange this to solve for n. So, we have \(n = p(1-p)*(z/E)^2\). For a margin of error of 1% or 0.01, and using the previous values for p and z, we substitute these values into the formula and calculate the required sample size.