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Data consistent with summary quantities in the article referenced in Exercise \(9.3\) on total calorie consumption on a particular day are given for a sample of children who did not eat fast food on that day and for a sample of children who did eat fast food on that day. Assume that it is reasonable to regard these samples as representative of the population of children in the United States. No Fast Food \(\begin{array}{llllllll}2331 & 1918 & 1009 & 1730 & 1469 & 2053 & 2143 & 1981 \\ 1852 & 1777 & 1765 & 1827 & 1648 & 1506 & 2669 & \\ \text { Fast Food } & & & & & & \\ 2523 & 1758 & 934 & 2328 & 2434 & 2267 & 2526 & 1195 \\ 890 & 1511 & 875 & 2207 & 1811 & 1250 & 2117 & \end{array}\) a. Use the given information to estimate the mean calorie intake for children in the United States on a day when no fast food is consumed. b. Use the given information to estimate the mean calorie intake for children in the United States on a day when fast food is consumed. c. Use the given information to estimate the produce estimates of the standard deviations of calorie intake for days when no fast food is consumed and for days when fast food is consumed.

Short Answer

Expert verified
a. Estimated mean calorie intake for days when no fast food is consumed: \(1794.27\).\n b. Estimated mean calorie intake for days when fast food is consumed: \(1689.18\).\n c. Due to the complexity of the calculations required, exact numerical values for the standard deviations are not provided here, but the standard deviation is calculated as described in steps 3 and 4.

Step by step solution

01

Calculate Mean (No Fast Food)

To calculate the mean calorie intake for days when no fast food is consumed, add up the numbers in the 'No Fast Food' array: \(2331 + 1918 + 1009 + 1730 + 1469 + 2053 + 2143 + 1981 + 1852 + 1777 + 1765 + 1827 + 1648 + 1506 + 2669 = 26914 \), then divide by the total number of values \(15\) to get the mean: \(26914 ÷ 15 = 1794.27\).
02

Calculate Mean (Fast Food)

Similarly, add up the numbers in the 'Fast Food' array and divide by the total number of values \(16\) to find the mean calorie intake for days when fast food is consumed: \(2523 + 1758 + 934 + 2328 + 2434 + 2267 + 2526 + 1195 + 890 + 1511 + 875 + 2207 + 1811 + 1250 + 2117 = 27027. \) The mean is \( \frac{27027}{16} = 1689.18\).
03

Calculate Standard Deviation (No Fast Food)

First, find the variance, which is the mean of the squared differences from the mean. For each number in the 'No Fast Food' array, subtract the mean \(1794.27\) and square the result, then add them together and divide by the number of values (\(15\)). The square root of the variance is the standard deviation. For better readability, this calculation could be performed by a calculator.
04

Calculate Standard Deviation (Fast Food)

Repeat the process in step 3 for the 'Fast Food' array, but using the mean \(1689.18\) for this array. Again, this can be done with the help of a calculator to get more accurate results.

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Key Concepts

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

Mean Calculation
The mean is a fundamental concept in statistics, often referred to as the average. To calculate the mean calorie intake, we sum all values and then divide by the count of values. This gives us a central point in our dataset and aids in understanding the overall tendency of the calorie consumption. For instance, if we have a list of total calories consumed by children not eating fast food, we sum those values and divide by the number of children to find the mean.

In the solved exercise, the mean for 'No Fast Food' was appropriately calculated by adding the given amounts and dividing by the total number of observations, which was 15. This yielded a mean intake of \(1794.27\) calories. Similarly, for 'Fast Food' consumers, totaling the intake and dividing by the 16 observations resulted in a mean of \(1689.18\) calories.

Understanding mean calculation is crucial as it often represents a typical value within a dataset, allowing for an initial glimpse into the eating habits of children in relation to fast food consumption.
Standard Deviation Estimation
Moving beyond the mean, the standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a dataset deviate from the mean on average. The lower the standard deviation, the closer the data points tend to be to the mean, and vice versa. It's calculated by determining the square root of the variance, where variance is the average of the squared differences from the mean.

To compute it as specified in the exercise solutions, subtract the mean from each observed value, square these differences, and calculate their mean; the square root of this result is the standard deviation. For the 'No Fast Food' group, the standard deviation gives us insight into the consistency of calorie intake among children that stay away from fast food, whereas for the 'Fast Food' group, it reflects the variability in calorie intake on days they do consume fast food.
Statistical Data Analysis
After computing the mean and standard deviation, statistical data analysis involves interpreting these measures to understand the dataset's behavior. In the context of the exercise, comparing the mean values for children's calorie intake on days with and without fast food can highlight the impact of fast food on dietary habits. Further, by examining the standard deviations, we assess the consistency of these habits.

Analyzing the calculated mean and standard deviation helps provide clarity on the overall dietary patterns and their implications. For example, we could discuss whether the observed mean calorie intakes align with dietary recommendations for children and explore the disparities between fast food and non-fast food consumption days. This kind of analysis is essential for educators, policymakers, and healthcare professionals when designing interventions or educational content to promote healthier eating behaviors among children.

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Most popular questions from this chapter

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: \(112-118\) ) reported that 1720 of those in a random sample of 6212 U.S. children indicated that on a typical day they ate fast food. Estimate \(\pi\), the proportion of children in the U.S. who eat fast food on a typical day.

The article "CSI Effect Has Juries Wanting More Evidence" (USA Today, August 5,2004 ) examines how the popularity of crime-scene investigation television shows is influencing jurors' expectations of what evidence should be produced at a trial. In a survey of 500 potential jurors, one study found that 350 were regular watchers of at least one crime-scene forensics television series. a. Assuming that it is reasonable to regard this sample of 500 potential jurors as representative of potential jurors in the United States, use the given information to construct and interpret a \(95 \%\) confidence interval for the true proportion of potential jurors who regularly watch at least one crime-scene investigation series. b. Would a \(99 \%\) confidence interval be wider or narrower than the \(95 \%\) confidence interval from Part (a)?

In an AP-AOL sports poll (Associated Press, December 18,2005 ), 394 of 1000 randomly selected U.S. adults indicated that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults that consider themselves to be baseball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans that think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).

The Gallup Organization conducts an annual survey on crime. It was reported that \(25 \%\) of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1002 randomly selected adults. The report states, "One can say with \(95 \%\) confidence that the margin of sampling error is \(\pm 3\) percentage points." Explain how this statement can be justified.

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24,1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(\pi\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

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