Question

10.48 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) at lunchtime in New York City were approached as they entered the restaurant 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, 3857 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 hamburger-chain 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 2000 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 3857 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 \(\alpha=.01\). 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. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a potential bias? Explain.

Short answer

The large standard deviation signifies high variability in calorie intake. Given the fact that calories consumed can't be negative, we can suggest data is not normally distributed. The results of a t-test on the sample provide evidence that the mean calories consumed surpass the recommended lunch intake of 750 calories. However, it's not unequivocally justifiable to extrapolate the results to all adult Americans. Using customer receipts ensures accurate data, whereas asking customers directly could introduce recall bias. Additionally, asking for a receipt before ordering could affect customer's behavior.

Step-by-step solution

Part a - Interpreting the Standard Deviation

The large value of standard deviation implies that there is a high variability in the amount of calories consumed by individual consumers. This means that while the calorie intake for most individuals might be closer to the mean (857 calories), some consumers are consuming significantly more or less calories than the mean.

Part b - Reasoning Why the Data is not Normally Distributed

Considering the fact that number of calories consumed can't be negative, and given a large standard deviation with a mean of 857, it is not reasonable to assume a normal distribution. This is because for a normal distribution, we would expect a significant proportion of data to fall within one standard deviation from the mean (i.e., between \(857 - 677 = 180\) and \(857 + 677 = 1534\) calories). But as calorie consumption less than zero is not possible, it is likely that the data is skewed to the right, making it not normally distributed.

Part c - Hypothesis Testing

Here, we're testing if the mean number of calories in a NYC hamburger-chain lunchtime purchase exceeds the recommended 750 calories. The null hypothesis \(H0\) is: the mean number of calories is less than or equal to 750, and the alternative hypothesis \(H1\) is: the mean number of calories is greater than 750. We will use a single-sample t-test due to large sample size and unknown population standard deviation.\n\nFirst, we need to calculate the test statistic, we use the formula \[ t = \frac{{\bar{x} - \mu}}{{s / \sqrt{n}}}\], where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean (which is our hypothesized value of 750 here), \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values into the formula yields t=(857-750)/(677/\sqrt{3857}).\n\nUsing statistical tables or an online calculator, we can then find the corresponding p-value for the calculated t statistic. If the p-value is less than the significance level stated (\(\alpha=.01\)), we reject the null hypothesis and conclude that the mean number of calories in a NYC hamburger-chain lunch time purchase is indeed greater than 750 calories.

Part d - Generalizing Results

Whether it would be reasonable to generalize the conclusions drawn from this study to all adult Americans would depend on the representativeness of the sample. If the sample is not representative of all adult Americans (due to factors like lifestyle, metabolism, food preferences or other geographical differences), conclusions drawn from this sample may not be applicable to all adult Americans.

Part e - Importance of Customer Receipt

Using a customer receipt to determine what was ordered provides an accurate and unbiased method of data collection. Conversely, asking a customer what they purchased could introduce recall bias as the customer may forget, understate or overstate what they actually ordered.

Part f - Potential Bias

Asking a customer to provide his or her receipt before they ordered could introduce a potential bias, as the awareness of being monitored could alter the customer's behavior and influence their ordering decision, they might order food with lower calories knowing their order is being recorded.

Key concepts

Sample Standard Deviation

When we talk about sample standard deviation, we're referring to a measure that reveals the extent of variation or dispersion from the average in a set of data. Imagine you and your friends measure how many steps you take daily. If everyone walks roughly the same number of steps each day, the standard deviation will be low. Conversely, if some friends barely walk while others are running marathons, the standard deviation will be high.

In our exercise, the large sample standard deviation tells us that diners at these New York City hamburger chains have a wide range in the number of calories they consume. While many might have meals with caloric content close to the sample mean of 857 calories, others might indulge in less calorie-dense options or more extravagant and calorie-heavy meals. This can pose challenges when trying to summarize the dietary behavior of the sample because the variability is significant.

Non-Normal Distribution

Many datasets in real life, including those concerning human behavior or biological measures, don't follow a perfect bell curve defined by normal distribution. Non-normal distribution arises when the data is skewed or has outliers that pull it away from being symmetrical.

With the caloric intake data from the fast-food study, it's expected that the data would not be normal because the number of calories one consumes cannot go below zero—nobody has a negative lunch! Additionally, the large sample standard deviation means the data points are spread out, making tails on the distribution graph that likely trail off towards the higher calorie counts. This right-skewed distribution means that most people consume fewer calories, with a smaller number tending to eat much more.

T-Test

A t-test is a statistical test used to compare the mean of a sample to a known value (in this case, the recommended calorie intake for lunch) or to compare the means of two samples. When we're dealing with a single sample and we don't know the population's standard deviation (which is common), we use a t-test instead of a z-test.

The t-test calculates a t-score, which helps us understand how far the sample mean deviates from the comparison value, considering the size of the sample and the variation within it. The larger the t-score, the more difference there is between the sample mean and the comparison value. When conducting the t-test for our New York City lunch example, using a significance level of 0.01 implies that we're very strict about claiming that something is not just due to chance—you need strong evidence to convince us that the average caloric intake is actually greater than the recommendation.

Caloric Intake Analysis

Analyzing caloric intake is crucial for understanding dietary habits and making recommendations for healthy eating. In the case of our fast-food study, we analyze the sample data to discern whether the average lunchtime caloric intake at hamburger chains exceeds the recommended 750 calories.

It’s essential to ensure that the methods used for data collection are accurate and unbiased. As noted in the exercise, using customer receipts is preferable to relying on self-reported data, as it mitigates issues like memory errors or intentional underreporting. Moreover, the analysis must be mindful of potential biases. For instance, if customers knew their receipts were going to be checked for a study before ordering, they may have altered their choices, which might not reflect their typical behavior. Thus, understanding and accounting for these aspects is pivotal in providing reliable conclusions from the caloric intake analysis.