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Chapter 15: Problem 18

An investigation carried out to study purchasers of luxury automobiles reported data on a number of different attributes that might affect purchase decisions, including comfort, safety, styling, durability, and reliability ("Measuring Values Can Sharpen Segmentation in the Luxury Car Market," Journal of Advertising Research \([1995]: 9-\) 22). Here is summary information on the level of importance of speed, rated on a seven-point scale: $$ \begin{array}{lccc} \text { Type of Car } & \text { American } & \text { German } & \text { Japanese } \\ \hline \text { Sample size } & 58 & 38 & 59 \\ \text { Sample mean rating } & 3.05 & 2.87 & 2.67 \end{array} $$ In addition, \(\mathrm{SSE}=459.04\). Carry out a hypothesis test to determine if there is sufficient evidence to conclude that the mean importance rating of speed is not the same for owners of these three types of cars.

Short Answer

Expert verified
The task of carrying out the hypothesis test cannot be fully completed because the information required to calculate SSb (Sum of Squares between) is not provided in the problem.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) states that the mean importance rating of speed is the same for all types of cars i.e., \(\mu_{American} = \mu_{German} = \mu_{Japanese}\). The alternative hypothesis \(H_a\) states that at least one mean is different, i.e., the mean importance rating of speed is not the same for all types.
02

Calculate degrees of freedom

The degrees of freedom between the groups (dfb) is calculated as \(k - 1\), where \(k\) is the number of groups (here, \(k = 3\)). So, \(dfb = 3 - 1 = 2\). The degrees of freedom within the groups (dfw), also known as error degrees of freedom, is calculated as \(N - k\), where \(N\) is the total number of observations. So, \(dfw = (58 + 38 + 59) - 3 = 152\).
03

Calculate Mean Squares

Mean Square between groups (MSb) is given by the formula \(\mathrm{SSb} / dfb\) and Mean Square within groups (MSw) is given by the formula \(\mathrm{SSw} / dfw\). Given that SSE or SSw = 459.04, we get MSw = 459.04 / 152 = 3.02.
04

Execute Analysis of Variance (ANOVA) Test

The F-statistic can be calculated by taking the ratio of the mean square between groups (MSb) and the mean square within groups (MSw), i.e., \(F = MSb / MSw\). But to perform this calculation, SSb (Sum of Squares between) is required which is not given in the problem and cannot be calculated with the available information in the problem. So, the solution stops here.

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Key Concepts

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

ANOVA
Analysis of Variance (ANOVA) is a statistical method used to compare the means of three or more samples to see if at least one sample mean is significantly different from the others. It's particularly useful when dealing with multiple groups, as it helps assess whether any of the group means are statistically different from each other.

In the context of the exercise, ANOVA is used to assess whether the mean importance rating of speed differs among American, German, and Japanese luxury car owners. By analyzing the variance within each group against the variance between the different groups, ANOVA helps determine if any observed differences in sample means are due to actual differences in population means or simply due to random chance.

Each ANOVA test follows a series of steps, and a key component involves calculating the F-statistic, which is the ratio of the variance between the groups to the variance within the groups. The F-statistic follows the F-distribution, which is determined by degrees of freedom associated with the variance of the means between the groups (numerator) and the variance within the groups (denominator).
Mean importance rating
The mean importance rating represents the average value assigned by a sample of participants rating a particular attribute or feature. In studies like this, participants could be asked to rate how important speed is to them in a luxury car on a scale of one to seven, with one being 'not important at all' and seven being 'extremely important'.

The calculated mean provides a concise summary of the group's preferences or perceptions towards the attribute in question鈥攊n this case, speed. In the exercise, the mean importance rating for speed was assessed for owners of different types of luxury cars鈥擜merican, German, and Japanese. Understanding how these means compare is essential in determining whether marketing strategies or product features should be tailored to specific consumer preferences linked to the country of origin of the car brand.
Degrees of freedom
Degrees of freedom are an essential concept in statistics that refer to the number of independent values or quantities that can be assigned to a statistical distribution. In the realm of hypothesis testing and particularly in ANOVA, degrees of freedom are used to determine the F-distribution, which is critical in establishing the cut-off points for deciding whether to accept or reject the null hypothesis.

The degrees of freedom for between-group variability, often symbolized as dfB or dfbetween, are calculated by subtracting one from the number of groups (\( k - 1 \)). For within-group variability, dfW or dfwithin, they are found by subtracting the number of groups from the total number of observations (\( N - k \)).

In our exercise, the degrees of freedom play a pivotal role in establishing the F-distribution for the ANOVA test. With the correct degrees of freedom, researchers can determine the critical value of the F-test and, thereby, decide if the evidence is strong enough to suggest significant differences in mean importance ratings of speed among the car owners grouped by the car's country of origin.

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Most popular questions from this chapter

Research carried out to investigate the relationship between smoking status of workers and short-term absenteeism rate (hr/mo) yielded the accompanying summary information ("Work-Related Consequences of Smoking Cessation," Academy of Management Journal [1989]: \(606-621) .\) In addition, \(F=2.56\). Construct an ANOVA table, and then state and test the appropriate hypotheses using a .01 significance level. $$\begin{array}{lrl} \text { Status } & \begin{array}{l} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{l} \text { Sample } \\ \text { Mean } \end{array} \\ \hline \text { Continuous smoker } & 96 & 2.15 \\ \text { Recent ex-smoker } & 34 & 2.21 \\ \text { Long-term ex-smoker } & 86 & 1.47 \\ \text { Never smoked } & 206 & 1.69 \end{array}$$

Give as much information as you can about the \(P\) -value for an upper-tailed \(F\) test in each of the following situations. a. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=5.37\) b. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=1.90\) c. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=15, F=4.89\) d. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=14.48\) e. \(\mathrm{df}_{1}=3, \mathrm{df}_{2}=20, F=2.69\) f. \(\mathrm{df}_{1}=4, \mathrm{df}_{2}=50, F=3.24\)

Suppose that a random sample of size \(n=5\) was selected from the vineyard properties for sale in Sonoma County, California, in each of three years. The following data are consistent with summary information on price per acre (in dollars, rounded to the nearest thousand) for disease-resistant grape vineyards in Sonoma County (Wines and Vines, November 1999). $$\begin{array}{llllll} 1996 & 30,000 & 34,000 & 36,000 & 38,000 & 40,000 \\ 1997 & 30,000 & 35,000 & 37,000 & 38,000 & 40,000 \\ 1998 & 40,000 & 41,000 & 43,000 & 44,000 & 50,000 \end{array}$$ a. Construct boxplots for each of the three years on a common axis, and label each by year. Comment on the similarities and differences. b. Carry out an ANOVA to determine whether there is evidence to support the claim that the mean price per acre for vineyard land in Sonoma County was not the same for the three years considered. Use a significance level of \(.05\) for your test.

The article "Utilizing Feedback and Goal Setting to Increase Performance Skills of Managers" (Academy of Management Journal \([1979]: 516-526\) ) reported the results of an experiment to compare three different interviewing techniques for employee evaluations. One method allowed the employee being evaluated to discuss previous evaluations, the second involved setting goals for the employee, and the third did not allow either feedback or goal setting. After the interviews were concluded, the evaluated employee was asked to indicate how satisfied he or she was with the interview. (A numerical scale was used to quantify level of satisfaction.) The authors used ANOVA to compare the three interview techniques. An \(F\) statistic value of \(4.12\) was reported. a. Suppose that a total of 33 subjects were used, with each technique applied to 11 of them. Use this information to conduct a level \(.05\) test of the null hypothesis of no difference in mean satisfaction level for the three interview techniques. b. The actual number of subjects on which each technique was used was \(45 .\) After studying the \(F\) table, explain why the conclusion in Part (a) still holds.

The nutritional quality of shrubs commonly used for feed by rabbits was the focus of a study summarized in the article "Estimation of Browse by Size Classes for Snowshoe Hare" (Journal of Wildlife Management [1980]: 34-40). The energy content (cal/g) of three sizes (4 mm or less, \(5-7 \mathrm{~mm}\), and \(8-10 \mathrm{~mm}\) ) of serviceberries was studied. Let \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) denote the true energy content for the three size classes. Suppose that \(95 \%\) simultaneous confidence intervals for \(\mu_{1}-\mu_{2}, \mu_{1}-\mu_{3}\), and \(\mu_{2}-\mu_{3}\) are \((-10,290),(150,450)\), and \((10,310)\), respectively. How would you interpret these intervals?
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