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A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]:1369- 1374). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) in New York City were approached as they entered the restaurant at lunchtime and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3,857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburgerchain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2,000 calories, the online Healthy Dining Finder (www.healthydiningfinder .com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3,857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=0.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased.

Short Answer

Expert verified
a. High standard deviation indicates high variance in lunchtime caloric intake among customers. b. As caloric intake can't be negative, it's unreasonable to expect a normal distribution. c. Conduct a one-sample t-test and compare t-score with t-critical value. If t-score>t-critical value, reject null hypothesis in favor of the alternative. d. One can't generalize the result to all U.S. adults because of varying eating habits, regional preferences, etc. e. Receipts give a more accurate account of the actual purchase compared to self-reports

Step by step solution

01

Understanding Standard Deviation

The standard deviation is a measure that gauges the amount of variation or dispersion of a set of values. If it's noticeably large, it indicates that the data points are spread out over a wider range of values. From the data, a large standard deviation signifies that the caloric intake of customers buying fast food from these restaurants at lunchtime varies a lot.
02

Distribution of Calories

Given the limited possibility of negative calories, a normal distribution theory might not be reasonable. Normal distribution is symmetrical, suggesting an equal chance of values which fall either side of the mean. However, calorie consumption can't be negative, thus the distribution wouldn't be symmetrical for caloric intake.
03

Hypothesis Testing

The null hypothesis is that the mean caloric intake is equal to the recommendation (750 calories) while the alternative hypothesis is that the mean caloric intake is greater than the recommendation. Using the one-sample t-test formula: \(t = \frac{857-750}{677/\sqrt{3857}}\). This value is then compared to the t-critical value for a one-tailed test with \(\alpha=0.01\) and degrees of freedom \(3857-1\). If the calculated t-score exceeds the critical value, then reject the null hypothesis in favor of the alternative.
04

Generalizing Conclusion

When it comes to generalizing the conclusion to lunchtime fast-food purchases of all American adults, one must consider how well the customers from the study represent all American adults. Factors such as different eating habits, regional food preferences, varying health conditions could influence the results, thus it may not be reasonable to generalize without a more diverse and larger sample size.
05

Data Collection Method

Using customer receipts provides a more accurate record of what was purchased. The receipt data is more reliable than self-reported data since customers might forget, give inaccurate accounts, or choose to hide their actual purchase.

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Key Concepts

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

Sample Standard Deviation
Understanding the 'sample standard deviation' is key when analyzing data because it describes how much variation exists in a dataset. In the context of the fast-food intake study, a large standard deviation of 677 calories suggests that the amount of calories New Yorkers consume at hamburger chains varies widely. Some customers may consume just a small salad, while others go for a large combo meal, leading to this broad spread in caloric intake.

It's vital to grasp that the standard deviation helps us anticipate the expected range of calories consumed. If we imagine plotting all the calorie counts, a large standard deviation implies that they are not clustered closely around the mean (857 calories); instead, they're stretched out across the calorie spectrum. In practice, this makes predicting an individual's caloric intake more challenging, since we'd expect a lot of different values, not just values around an 'average' meal.
Normal Distribution Theory
The 'normal distribution theory' is a cornerstone of statistics, often used to model natural occurrences and understand probabilities. It's recognizable by its symmetrical, bell-shaped curve, where most data points cluster around the mean and fewer occur as you move away. However, this works best when the data can theoretically take any value, including negative ones.

In our fast-food scenario, the theory hits a snag – calories cannot be negative. This limitation introduces a skew in the data; imagine trying to fit a symmetrical bell curve to a dataset that bunches up on one side because it can't extend below zero. Hence, expecting caloric intake to follow a normal distribution is unrealistic in this study, affecting the analysis and the types of statistical tests we can apply reliably. For phenomena with natural boundaries, like caloric content, we must look to other distribution models that better fit our dataset.
One-Sample T-Test
The 'one-sample t-test' is a statistical procedure we employ to figure out if there's a significant difference between a sample mean and a known or hypothesized value of the population mean. In the fast-food study, this test helps us assess whether New York City's hamburger-chain lunchtime purchase exceeds the proposed 750-calorie lunch target.

To conduct this test, we set up two hypotheses: the null, which states no difference (mean equals 750 calories), and the alternative, which suggests a higher value. The calculated t-value from our formula will then inform us if we can reject the null hypothesis in favor of the alternative at the given significance level \(\alpha = 0.01\). A significant result means the data provides strong evidence that citizens are indeed consuming more calories than recommended. However, this only speaks to the sample — extrapolating these results to the larger population requires careful consideration of the sample's representativeness.

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Most popular questions from this chapter

The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llllllll}3.80 & 5.89 & 4.92 & 3.88 & 2.65 & 5.57 & 6.39 & 3.24\end{array}\) The mean price of a Big Mac in the U.S. in May 2009 was \$3.57. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12)

Because of safety considerations, in May, \(2003,\) the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage) in warm months and 185 pounds as a typical weight in cold months. The Alaska Journal of Commerce (May 25,2003\()\) reported that Frontier Airlines conducted a study to estimate mean passenger plus carry-on weights. They found an mean summer weight of 183 pounds and a winter mean of 190 pounds. Suppose that these estimates were based on random samples of 100 passengers and that the sample standard deviations were 20 pounds for the summer weights and 23 pounds for the winter weights. a. Construct and interpret a \(95 \%\) confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a \(95 \%\) confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9,2005 ). The sample mean was 12.7 hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours. c. Explain why the null hypothesis was rejected in the test of Part (b) but not in the test of Part (a).

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? a. \(10 \mathrm{df}\), between -2.23 and 2.23 b. 24 df, between -2.80 and 2.80 c. \(24 \mathrm{df}\), to the right of 2.80

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