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Chapter 11: Problem 8

To investigate the real cost of owning different makes and models of new automobiles, a consumer protection agency followed 16 owners of new vehicles of four popular makes and models, call them \(T C, H A, N A,\) and \(F T,\) and kept a record of each of the owner's real cost in dollars for the first five years. The five-year costs of the 16 car owners are given below: $$ \begin{array}{|l|l|l|c|} \hline \mathrm{TC} & \mathrm{HA} & \mathrm{NA} & \mathrm{FT} \\ \hline 8423 & 7776 & 8907 & 10333 \\ \hline 7889 & 7211 & 9077 & 9217 \\ \hline 8665 & 6870 & 8732 & 10540 \\ \hline & 7129 & 9747 & \\ \hline & 7359 & 8677 & \\ \hline \end{array} $$ Test, using the ANOVA F-test at the 5\% level of significance, whether the data provide sufficient evidence to conclude that there are differences among the mean real costs of ownership for these four models.

Short Answer

Expert verified
There is sufficient evidence to conclude that there are differences in the mean costs among the models.

Step by step solution

01

State the Hypotheses

Begin by stating the null and alternative hypotheses for the ANOVA test. - Null Hypothesis (H_0): There is no difference in the mean real costs among the four car models (H_0: \mu_{TC} = \mu_{HA} = \mu_{NA} = \mu_{FT}).- Alternative Hypothesis (H_a): At least one model's mean cost is different (H_a: At least one \(\mu_k\) differs).
02

Calculate the Group Means and Overall Mean

Calculate the mean real cost for each car model (TC, HA, NA, FT) and the overall mean of all the groups combined. - For TC: Mean = \(\frac{8423 + 7889 + 8665}{3}\)- For HA: Mean = \(\frac{7776 + 7211 + 6870 + 7129 + 7359}{5}\)- For NA: Mean = \(\frac{8907 + 9077 + 8732 + 9747 + 8677}{5}\)- For FT: Mean = \(\frac{10333 + 9217 + 10540}{3}\)
03

Compute the ANOVA Components

Calculate the Sum of Squares Between Groups (SSB) and Within Groups (SSW). Use these to calculate the Mean Square Between Groups (MSB) and Mean Square Within Groups (MSW). - SSB and MSB will require using each group's mean and the overall mean. - SSW and MSW are obtained from each data point's deviation from its group mean.
04

Calculate the F-statistic

Using the formulas \(MSB = \frac{SSB}{df_{between}}\) and \(MSW = \frac{SSW}{df_{within}}\), compute the F-statistic: \(F = \frac{MSB}{MSW}\).
05

Determine the Critical Value and Compare

Determine the critical F-value for \(\alpha = 0.05\) with the degrees of freedom (df) calculated: df between = 3, df within = (total sample size - number of groups). Compare the calculated F-statistic with the critical value to decide if H_0 is rejected.
06

Conclusion

If the F-statistic > F-critical, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental concept in statistics. In the context of an ANOVA F-test, the null hypothesis often posits that there are no significant differences among the groups being compared. For this particular exercise, the null hypothesis is denoted as \( H_0: \mu_{TC} = \mu_{HA} = \mu_{NA} = \mu_{FT} \). This means that we assume all four car models, TC, HA, NA, and FT, have the same average real costs of ownership, over the first five years.

If the null hypothesis is not rejected, it suggests that any observed differences in sample means are due to random chance. However, rejecting the null hypothesis would imply that at least one car model has a significantly different mean real cost than the others, thus indicating a true difference in real costs among the groups.
Mean Real Costs
The mean real cost is an average value calculated from the data collected. It provides a central value around which each group's data points cluster. To find the mean for each car model, sum up all real costs for that model and divide by the number of data points.

For example, the mean for the TC model is calculated as \( \frac{8423 + 7889 + 8665}{3} \), while for HA, it is \( \frac{7776 + 7211 + 6870 + 7129 + 7359}{5} \). These means are essential in comparing whether real costs differ significantly across the groups.

Each mean represents the average ownership cost for a particular model. By comparing these means, we can understand the variability and differences in real costs between the car models.
Critical F-value
The critical F-value is a threshold value that the computed F-statistic is compared against to determine statistical significance in an ANOVA test. It depends on the selected significance level, \( \alpha \), and the degrees of freedom for between-group and within-group variations.

In this exercise, the critical F-value is determined for a significance level of \( \alpha = 0.05 \). This implies a 5% risk of concluding that a difference exists when there is none. The degrees of freedom needed to find this value in the F-distribution table are df between, which is the number of groups minus one, and df within, which is the total sample size minus the number of groups.

By comparing the F-statistic with the critical F-value, we can decide whether to reject the null hypothesis or not. If the F-statistic exceeds the critical F-value, this suggests strong evidence against the null hypothesis.
Degrees of Freedom
Degrees of freedom are crucial in the context of statistical tests like ANOVA. They help determine the variability in data and influence the critical F-value. For ANOVA, degrees of freedom are split into two components: between groups and within groups.

The degrees of freedom for between groups, denoted as \( df_{between} \), is calculated by subtracting one from the number of groups. In the given exercise, with four groups (car models), \( df_{between} = 4 - 1 = 3 \).

The degrees of freedom within groups, \( df_{within} \), are found by subtracting the number of groups from the total number of observations. If we assume that the total sample size is the sum of all individual observations, then \( df_{within} = 16 - 4 = 12 \).

These degrees of freedom are vital when looking up critical values in the F-distribution table and contribute significantly to calculating the ANOVA components like MSB and MSW.

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Most popular questions from this chapter

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