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Chapter 3: Problem 28

Predicting cost of meal from rating Refer to the previous exercise. The correlation with the cost of a dinner is 0.68 for food quality rating, 0.69 for service rating, and 0.56 for d茅cor rating. According to the definition of \(r^{2}\) as a measure for the reduction in the prediction error, which of these three ratings can be used to make the most accurate predictions for the cost of a dinner: quality of food, service, or d茅cor? Why?

Short Answer

Expert verified
Service rating provides the most accurate predictions, with an R-squared value of 0.4761.

Step by step solution

01

Understanding Correlation Values

The correlation coefficient measures the strength and direction of a linear relationship between two variables. The closer the value is to 1 or -1, the stronger the linear relationship. Here, we analyze three correlation coefficients: food quality rating (0.68), service rating (0.69), and d茅cor rating (0.56).
02

Calculating R-squared Values

The R-squared value is found by squaring the correlation coefficient: \(r^2 = r \times r\). This value represents the proportion of variance in the dependent variable (cost of dinner) that can be explained by the independent variable (rating). Calculate:- Food quality rating: \(0.68^2 = 0.4624\)- Service rating: \(0.69^2 = 0.4761\)- D茅cor rating: \(0.56^2 = 0.3136\).
03

Comparison of R-squared Values

Compare the R-squared values to determine which rating most accurately predicts the cost of a dinner:- Food quality rating (\(0.4624\))- Service rating (\(0.4761\))- D茅cor rating (\(0.3136\)). Service rating has the highest R-squared value at 0.4761, meaning it explains the most variance in dinner cost.
04

Conclusion

Because the service rating has the highest R-squared value (0.4761), it provides the most accurate prediction of the cost of a dinner based on the given ratings. The higher R-squared value indicates a better fit for the model of predicting cost from service rating.

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Key Concepts

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

R-squared
R-squared is a statistical measure that explains how much of the variability in the response variable can be explained by the predictor variable(s). In the context of predicting the cost of a dinner, R-squared helps us understand how well a particular rating can account for variations in dinner prices.
For each rating (food quality, service, and d茅cor), we calculate R-squared by squaring their respective correlation coefficients:
  • Food quality: The correlation is 0.68, giving an R-squared of 0.4624.
  • Service: The correlation is 0.69, producing an R-squared of 0.4761.
  • D茅cor: The correlation is 0.56, resulting in an R-squared of 0.3136.
These values show us how much of the cost variation each rating explains. For example, a higher R-squared value, such as the one for service, indicates a more reliable prediction model.
Prediction Error
Prediction error is the discrepancy between the actual value and the predicted one in a model. The aim of R-squared is to minimize this error by quantifying how much of the prediction can be made accurately.
When using service rating to predict dinner costs, the R-squared value is 0.4761. This means 47.61% of the variability in dinner costs is accounted for by the service rating. The remaining variability is what we consider the prediction error.
  • The lower the prediction error, the better the model predicts.
  • A high R-squared value corresponds to a smaller prediction error.
In our comparison, service rating gives the smallest prediction error, making it the most accurate predictor for dinner cost.
Linear Relationship
A linear relationship is characterized by a straight line when graphically representing two variables. The correlation coefficients give us an insight into how strongly one variable predicts another based on this linear association.
In our exercise, we have three correlation coefficients:
  • Food quality: 0.68
  • Service: 0.69
  • D茅cor: 0.56
The strength of these coefficients shows a positive linear relationship, meaning as the rating increases, the cost of dinner likely increases as well. A higher correlation value indicates a stronger linear relationship, which is crucial for making accurate predictions. The service rating, with the highest correlation of 0.69, suggests the strongest linear relationship with dinner cost.
Regression Analysis
Regression analysis is a statistical tool used to understand the relationship between different variables. In our case, it's used to see how well different ratings can predict the cost of a dinner.
To conduct a simple regression analysis, we start by determining the correlation between each rating and dinner cost, then move to calculate the R-squared for each. This tells us which rating most significantly impacts dinner cost.
The steps include:
  • Identifying correlations between variables.
  • Calculating R-squared for each correlation to evaluate prediction reliability.
  • Comparing R-squared values to find the best predictor.
The service rating, with the highest correlation and R-squared, shows a clear regression result: it is the best among the tested ratings for accurately predicting dinner costs.

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Most popular questions from this chapter

NL baseball Example 9 related \(y=\) team scoring (per game) and \(x=\) team batting average for American League teams. For National League teams in 2010 , \(\hat{y}=-6.25+41.5 x\). (Data available on the book's website in the NL team statistics file.) a. The team batting averages fell between 0.242 and \(0.272 .\) Explain how to interpret the slope in context. b. The standard deviations were 0.00782 for team batting average and 0.3604 for team scoring. The correlation between these variables was 0.900 . Show how the correlation and slope of 41.5 relate in terms of these standard deviations. c. Software reports \(r^{2}=0.81 .\) Explain how to interpret this measure.

According to data obtained from the General Social Survey (GSS) in 2014,1644 out of 2532 respondents were female and interviewed in person, 551 were male and interviewed in person, 320 were female and interviewed over the phone and 17 were male and interviewed over the phone. a. Explain how we could regard either variable (gender of respondent, interview type) as a response variable. b. Display the data as a contingency table, labeling the variables and the categories. c. Find the conditional proportions that treat interview type as the response variable and gender as the explanatory variable. Interpret. d. Find the conditional proportions that treat gender as the response variable and interview type as the explanatory variable. Interpret. e. Find the marginal proportion of respondents who (i) are female, (ii) were interviewed in person.

For the 100 cars on the lot of a used-car dealership, would you expect a positive association, negative association, or no association between each of the following pairs of variables? Explain why. a. The age of the car and the number of miles on the odometer b. The age of the car and the resale value c. The age of the car and the total amount that has been spent on repairs d. The weight of the car and the number of miles it travels on a gallon of gas e. The weight of the car and the number of liters it uses per \(100 \mathrm{~km}\).

According to data selected from GSS in \(2014,\) the correlation between \(y=\) email hours per week and \(x=\) ideal number of children is -0.0008 a. Would you call this association strong or weak? Explain. b. The correlation between email hours per week and Internet hours per week is \(0.33 .\) For this sample, which explanatory variable, ideal number of children or Internet hours per week, seems to have a stronger association with \(y\) ? Explain.

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