Question

The average American has become accustomed to eating away from home, especially at fast-food restaurants. Partly as a result of this fast-food habit, the per-capita consumption of cheese (the main ingredient in pizza) and nondiet soft drinks has risen dramatically from a decade ago. A study in American Demographics reports that the average American consumes 25.7 pounds of cheese and drinks 40 gallons (or approximately 645 8-ounce servings) of nondiet soft drinks per year. \({ }^{17}\) To test the accuracy of these reported averages, a random sample of 40 consumers is selected, and these summary statistics are recorded: $$\begin{array}{lcc} & \text { Cheese (lb/yr) } & \text { Soft Drinks (gal/yr) } \\\\\hline \text { Sample Mean } & 28.1 & 39.2 \\\\\text { Sample Standard Deviation } & 3.8 & 4.5\end{array}$$ Use your knowledge of statistical estimation to estimate the average per- capita annual consumption for these two products. Does this sample cause you to support or to question the accuracy of the reported averages? Explain.

Short answer

Answer: We can question the accuracy of the reported average for cheese consumption, but we can support the accuracy of the reported average for soft drink consumption.

Step-by-step solution

Analyze given data

We have the following data: Sample size (n): 40 consumers Reported cheese consumption (mean): 25.7 lb/year Reported soft drink consumption (mean): 40 gallons/year Sample mean cheese consumption (x̄₁): 28.1 lb/year Sample mean soft drink consumption (x̄₂): 39.2 gallons/year Sample standard deviation cheese consumption (s₁): 3.8 lb/year Sample standard deviation soft drink consumption (s₂): 4.5 gallons/year

Calculate the standard error of the means

The standard error (SE) of the mean can be calculated using the following formula: SE = s / sqrt(n) For cheese, SE₁ = 3.8 / sqrt(40) ≈ 0.60 For soft drinks, SE₂ = 4.5 / sqrt(40) ≈ 0.71

Find the confidence intervals for the means

Now, we will determine the 95% confidence intervals for the means. We use the formula: CI = x̄ ± (t × SE) In this case, the t value for a 95% confidence interval with 39 degrees of freedom (n-1) is approximately 2.021 (from a t-distribution table). For cheese, CI₁ = 28.1 ± (2.021 × 0.60) ≈ (27.1, 29.1) lb/year For soft drinks, CI₂ = 39.2 ± (2.021 × 0.71) ≈ (38.0, 40.4) gallons/year

Compare the confidence intervals with the reported averages

Now, we will compare our confidence intervals with the reported averages to see if the reported averages are within the 95% confidence intervals of the sample means. For cheese: The reported average of 25.7 lb/year does not fall within the confidence interval of (27.1, 29.1) lb/year calculated for the sample mean. Thus, we can question the accuracy of the reported average for cheese consumption. For soft drinks: The reported average of 40 gallons/year falls within the confidence interval of (38.0, 40.4) gallons/year calculated for the sample mean. This suggests that we can support the accuracy of the reported average for soft drink consumption. In conclusion, based on the given sample data, we can question the accuracy of the reported average for cheese consumption but support the accuracy of the reported average for soft drink consumption.

Key concepts

Confidence Intervals

When discussing statistical estimation, confidence intervals are crucial. They provide a range of values which can be used to estimate an unknown population parameter with a certain degree of confidence. For instance, in estimating the average consumption of cheese and soft drinks, confidence intervals give us a range wherein the true average likely lies.

Confidence intervals are calculated using the sample mean, standard error, and a t-value derived from the t-distribution table, reflecting the desired level of certainty, such as 95% in our case. Mathematically, the formula is:\[ CI = \bar{x} \pm (t \times SE)\]

• For cheese consumption: The confidence interval was calculated as \((27.1, 29.1)\) pounds/year.
• For soft drink consumption: The confidence interval was \((38.0, 40.4)\) gallons/year.

Ultimately, confidence intervals help in determining whether the observed data supports or contradicts a prior claim about a population parameter, such as average consumption in this case.

Sample Mean

The sample mean, denoted as \(\bar{x}\), is a critical measure in statistics, representing the average value of a sample. When scientists and researchers conduct studies, they often work with samples rather than entire populations. Therefore, understanding the sample mean is vital to interpret results accurately.

In our exercise, the sample mean for cheese consumption was found to be 28.1 lb/year, while that for soft drinks was 39.2 gallons/year. These figures contrast with the reported population means of 25.7 lb/year and 40 gallons/year, respectively.

Using sample means, we can construct confidence intervals and perform numerous statistical tests to make inferences about broader population behaviors. The sample mean is often the starting point in these statistical endeavors.

Standard Error

Standard error (SE) quantifies the accuracy with which a sample mean represents the entire population. It is affected by two elements: the sample's standard deviation and the size of the sample.

The formula for standard error is given by: \[ SE = \frac{s}{\sqrt{n}} \] where:\

• \(s\) is the sample standard deviation, and
• \(n\) is the sample size.

In this problem, the standard error for cheese consumption was calculated to be approximately 0.60, and for soft drinks, it was about 0.71.

The smaller the standard error, the more representative the sample mean is of the population mean. It's a crucial component in constructing confidence intervals that provide insights into how much the sample mean can deviate from the population mean.