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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

Expert verified
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

01

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.
02

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.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values that is likely to contain an unknown population parameter with a given level of confidence. In the context of the given exercise, we aim to construct a 95% confidence interval for the proportion of U.S. adults who favor 'sin taxes' on soda and junk food. The key here is to understand that a confidence interval provides a spectrum of possible values, not a definitive answer.
  • To calculate a confidence interval for a proportion, we use the formula: \[ p \pm z \left( \sqrt{\frac{p(1-p)}{n}} \right) \] where \(p\) is the sample proportion, \(z\) is the z-score corresponding to the desired confidence level, and \(n\) is the sample size.
  • For a 95% confidence level, the z-score is typically 1.96, which represents the number of standard deviations away from the mean that encompasses 95% of the data.
By using this formula, we can determine a range around our sample proportion (32%) where the true population proportion is likely to fall. It's essential to remember that the confidence interval itself does not imply that 95% of the population falls within that range, but rather, 95% of intervals calculated in this way would contain the true population proportion.
Margin of Error
The margin of error is a critical metric in statistics that shows the extent to which you can expect your sample estimates to deviate from the true population value. In the exercise example, the margin of error helps inform us of the precision of our estimate of adults who favor 'sin taxes'.
  • It is derived from the variability inherent in sample data and the chosen confidence level.
  • The margin of error can be calculated by subtracting the lower limit of the confidence interval from the sample proportion (if you prefer to think in terms of the difference from the estimate, use the upper and lower limits to find it from mid-point estimates).
  • Using the confidence interval formula: \[ \text{Margin of Error} = z \left( \sqrt{\frac{p(1-p)}{n}} \right) \]
High margins of error indicate less confidence in your sample result's precision, whereas a smaller margin suggests more precise estimates. Therefore, understanding and minimizing the margin of error is key in designing surveys and interpreting results.
Sample Size Calculation
Sample size calculation ensures that the survey results have a desired level of precision and confidence. In the scenario provided, we want a margin of error of only 1% with 95% confidence, and this requires an adequate sample size.
  • The formula used to determine the sample size required for a certain margin of error is: \[ n = \frac{p(1-p) (z/E)^2} \] where \(E\) is the desired margin of error, \(p\) is the sample proportion, and \(z\) is the z-score for the confidence level.
  • This formula rearranges the margin of error equation to solve for \(n\), highlighting that as the margin of error or the width of confidence interval becomes smaller, the needed sample size increases significantly.
  • By plugging the values of \(p = 0.32\), \(z = 1.96\), and \(E = 0.01\) (1%), you can calculate the sample size ensuring the survey's results are as precise as required. This consideration is crucial when planning surveys for statistical reliability without unnecessary expenditure of resources.
Understanding sample size calculation is critical for ensuring that survey findings are generalizable and reliable within acceptable limits of error.

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Most popular questions from this chapter

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\).

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To measure the effectiveness of a new teaching method for math in elementary school, each student in a class getting the new instructional method is matched with a student in a separate class on \(\mathrm{IQ}\), family income, math ability level the previous year, reading level, and all demographic characteristics. At the end of the year, math ability levels are measured again.

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=10.1, s_{1}=2.3, n_{1}=50\) and \(\bar{x}_{2}=12.4, s_{2}=5.7, n_{2}=50 .\)

We examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within ±3 , if the estimated standard deviation is \(\tilde{\sigma}=100\). If the estimated standard deviation is \(\tilde{\sigma}=50\). If the estimated standard deviation is \(\tilde{\sigma}=10 .\) Comment on how the variability in the population influences the sample size needed to reach a desired level of accuracy.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9, s_{d}=\) \(143.6, n_{d}=100\)

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