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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

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

Statistical Inference in Babies Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study, \(1 \quad 8\) -month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an "expected" population, with balls in the same color proportions as the sample, while other boxes had an "unexpected" population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds \((\mathrm{sd}=4.5\) seconds) and the expected populations for an average of 7.5 seconds \((\mathrm{sd}=4.2\) seconds). The sample size in each group was \(20,\) and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? (a) State the null and alternative hypotheses. (b) Calculate the relevant sample statistic. (c) Calculate the t-statistic.

Use a t-distribution to answer the question. Assume the sample is a random sample from a distribution that is reason ably normally distributed and we are doing inference for a sample mean. Find the area in a t-distribution above 1.5 if the sample has size \(n=8\).

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

Do Babies Understand Probability? Can babies reason probabilistically? A study \(^{19}\) investigates this by showing ten- to twelve-month-old infants two jars of lollipop-shaped objects colored pink or black. Each infant first crawled or walked to whichever color they wanted, determining their "preferred" color. They were then given the choice between two jars that had the same number of preferred objects, but that differed in their probability of getting the preferred color; each jar had 12 in the preferred color and either 4 or 36 in the other color. Babies choosing randomly or based on the absolute number of their preferred color would choose equally between the two jars, while babies understanding probability would more often choose the jar with the higher proportion of their preferred color. Of the 24 infants studied, 18 chose the jar with the higher proportion of their preferred color. Are infants more likely to choose the jar with the higher proportion of their preferred color? (a) State the null and alternative hypotheses. (b) Give the relevant sample statistic, using correct notation. (c) Which of the following should be used to calculate a p-value for this dataset? A randomization test, a test using the normal distribution, or either one? Why? (d) Find a p-value using a method appropriate for this data situation. (e) Make a conclusion in context, using \(\alpha=0.05\).

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