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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—in this case, speed. In the exercise, the mean importance rating for speed was assessed for owners of different types of luxury cars—American, 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

The article "Growth Response in Radish to Sequential and Simultaneous Exposures of \(\mathrm{NO}_{2}\) and \(\mathrm{SO}_{2} "(\) Environmental Pollution \([1984]: 303-325\) ) compared a control group (no exposure), a sequential exposure group (plants exposed to one pollutant followed by exposure to the second four weeks later), and a simultaneous-exposure group (plants exposed to both pollutants at the same time). The article states, "Sequential exposure to the two pollutants had no effect on growth compared to the control. Simultaneous exposure to the gases significantly reduced plant growth." Let \(\bar{x}_{1}, \bar{x}_{2}\), and \(\bar{x}_{3}\) represent the sample means for the control, sequential, and simultaneous groups, respectively. Suppose that \(\bar{x}_{1}>\bar{x}_{2}>\bar{x}_{3}\). Use the given information to construct a table where the sample means are listed in increasing order, with those that are judged not to be significantly different underscored. \(15.30\) 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 \((\mathrm{cal} / \mathrm{g})\) of three sizes \((4 \mathrm{~mm}\) 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?

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. $$\begin{array}{lrl} \text { Status } & \begin{array}{l} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{l} \text { Sample } \\ \text { Mean } \end{array} \\ \hline \text { Continuous smoker } & 96 & 2.15 \\ \text { Recent ex-smoker } & 34 & 2.21 \\ \text { Long-term ex-smoker } & 86 & 1.47 \\ \text { Never smoked } & 206 & 1.69 \end{array}$$

It has been reported that varying work schedules can lead to a variety of health problems for workers. The article "Nutrient Intake in Day Workers and Shift Workers" (Work and Stress [1994]: 332-342) reported on blood glucose levels (mmol/L) for day-shift workers and workers on two different types of rotating shifts. The sample sizes were \(n_{1}=37\) for the day shift, \(n_{2}=34\) for the second shift, and \(n_{3}=25\) for the third shift. A single- factor ANOVA resulted in \(F=3.834\). At a significance level of .05, does true average blood glucose level appear to depend on the type of shift?

The article "The Soundtrack of Recklessness: Musical Preferences and Reckless Behavior Among Adolescents" (Journal of Adolescent Research [1992]: \(313-331\) ) described a study whose purpose was to determine whether adolescents who preferred certain types of music reported higher rates of reckless behaviors, such as speeding, drug use, shoplifting, and unprotected sex. Independently chosen random samples were selected from each of four groups of students with different musical preferences at a large high school: (1) acoustic/pop, (2) mainstream rock, $$\begin{array}{ccccccccc} \text { Type of Box } & & & {\text { Compression Strength (Ib) }} & & & & {\text { Sample Mean }} & {\text { Sample SD }} \\ \hline 1 & 655.5 & 788.3 & 734.3 & 721.4 & 679.1 & 699.4 & 713.00 & 46.55 \\ 2 & 789.2 & 772.5 & 786.9 & 686.1 & 732.1 & 774.8 & 756.93 & 40.34 \\ 3 & 737.1 & 639.0 & 696.3 & 671.7 & 717.2 & 727.1 & 698.07 & 37.20 \\ 4 & 535.1 & 628.7 & 542.4 & 559.0 & 586.9 & 520.0 & 562.02 & 39.87 \\ & & & & & & & \overline{\bar{x}} =682.50 & \end{array}$$ (3) hard rock, and (4) heavy metal. Each student in these samples was asked how many times he or she had engaged in various reckless activities during the last year. The following table lists data and summary quantities on driving over \(80 \mathrm{mph}\) that is consistent with summary quantities given in the article (the sample sizes in the article were much larger, but for the purposes of this exercise, we use \(\left.n_{1}=n_{2}=n_{3}=n_{4}=20\right)\) $$\begin{array}{rrrr} \text { Acoustic/Pop } & \text { Mainstream Rock } & \text { Hard Rock } & \text { Heavy Metal } \\ \hline 2 & 3 & 3 & 4 \\ 3 & 2 & 4 & 3 \\ 4 & 1 & 3 & 4 \\ 1 & 2 & 1 & 3 \\ 3 & 3 & 2 & 3 \\ 3 & 4 & 1 & 3 \\ 3 & 3 & 4 & 3 \\ 3 & 2 & 2 & 3 \\ 2 & 4 & 2 & 2 \\ 2 & 4 & 2 & 4 \\ 1 & 4 & 3 & 4 \\ 3 & 4 & 3 & 5 \\ 2 & 2 & 4 & 4 \\ 2 & 3 & 3 & 5 \\ 2 & 2 & 3 & 3 \\ 3 & 2 & 2 & 4 \\ 2 & 2 & 3 & 5 \\ 2 & 3 & 4 & 4 \\ 3 & 1 & 2 & 2 \\ 4 & 3 & 4 & 3 \\ 20 & 20 & 20 & 20 \\ 2.50 & 2.70 & 2.75 & 3.55 \\ .827 & .979 & .967 & .887 \\ 6830 & 0584 & 0351 & 7868 \end{array}$$ Also, \(N=80\), grand total \(=230.0\), and \(\overline{\bar{x}}=230.0 / 80=\) 2.875. Carry out an \(F\) test to determine if these data provide convincing evidence that the true mean number of times driving over 80 mph varies with musical preference.

The accompanying data resulted from a flammability study in which specimens of five different fabrics were tested to determine burn times. $$\begin{array}{lllllll} & \mathbf{1} & 17.8 & 16.2 & 15.9 & 15.5 & \\ & \mathbf{2} & 13.2 & 10.4 & 11.3 & & \\ \text { Fabric } & \mathbf{3} & 11.8 & 11.0 & 9.2 & 10.0 & \\ & \mathbf{4} & 16.5 & 15.3 & 14.1 & 15.0 & 13.9 \\ & \mathbf{5} & 13.9 & 10.8 & 12.8 & 11.7 & \end{array}$$ \(\begin{aligned} \mathrm{MSTr} &=23.67 \\ \mathrm{MSE} &=1.39 \\ F &=17.08 \\ P \text { -value } &=.000 \end{aligned}\) The accompanying output gives the T-K intervals as calculated by MINITAB. Identify significant differences and give the underscoring pattern. $$\begin{array}{lrrr} & 1 & 2 & 3 & 4 \\ 2 & 1.938 & & & \\ & 7.495 & & & \\ 3 & 3.278 & -1.645 & & \\ & 8.422 & 3.912 & & \\ 4 & 3.830 & -0.670 & -2.020 & \\ & 1.478 & -3.445 & -4.372 & 0.220 \\ 5 & 6.622 & 2.112 & 0.772 & 5.100 \end{array}$$

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