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Chapter 15: Problem 22

The article "Heavy Drinking and Problems Among Wine Drinkers" (Journal of Studies on Alcohol [1999]: 467-471) analyzed drinking problems among Canadians. For each of several different groups of drinkers, the mean and standard deviation of "highest number of drinks consumed" were calculated: \(\bar{x}\) $$\begin{array}{lccc} & \overline{\boldsymbol{x}} & \boldsymbol{s} & {n} \\ \hline \text { Beer only } & 7.52 & 6.41 & 1256 \\ \text { Wine only } & 2.69 & 2.66 & 1107 \\ \text { Spirits only } & 5.51 & 6.44 & 759 \\ \text { Beer and wine } & 5.39 & 4.07 & 1334 \\ \text { Beer and spirits } & 9.16 & 7.38 & 1039 \\ \text { Wine and spirits } & 4.03 & 3.03 & 1057 \\ \text { Beer, wine, and spirits } & 6.75 & 5.49 & 2151 \end{array}$$ Assume that each of the seven samples studied can be viewed as a random sample for the respective group. Is there sufficient evidence to conclude that the mean value of highest number of drinks consumed is not the same for all seven groups?

Short Answer

Expert verified
Since the conclusion is based on the statistical analysis result which requires calculation, without computed data, it's not feasible to present a definitive conclusion here. The decision is made by comparing the computed F statistic with the critical F value from F-distribution chart. If F statistic > F critical, then null hypothesis is rejected implying there is at least one group mean that is different.

Step by step solution

01

Setup the hypothesis

The null hypothesis (\(H_0\)): The means are equal across all groups. The alternative hypothesis (\(H_1\)): At least one group mean is different.
02

Compute the Sum of Squares Within (SSW)

First, calculate the Sum of Squares Within (SSW) by using the formula: \[SSW = \sum_i^n s_i^2 * (n_i - 1)\] where \(s_i\) is the standard deviation and \(n_i\) is the sample size of ith group. SSW measures the variation of observations within each group.
03

Compute the Sum of Squares Between (SSB)

Then, compute Sum of Squares Between (SSB) by using the formula: \[SSB = \sum_i^n n_i * (x_i - x_total)^2\] where \(x_i\) is the mean of the ith group, \(x_total\) is the total mean which is computed as \[x_total = \frac{\sum_i^n (x_i * n_i)}{\sum_i^n n_i}\] and \(n_i\) is the sample size of the ith group. SSB measures the variation between the group means.
04

Calculate Degree of Freedom

Calculate Degree of Freedom Within (dfW) as: \[dfW = \sum_i^n (n_i - 1)\] And Degree of Freedom Between (dfB) as: \[dfB = n - 1\] where n is the number of groups.
05

Calculate the F-Statistic

Compute the F-Statistic using the formula: \[F = \frac{SSB / dfB}{SSW / dfW}\]
06

Make a Conclusion

Compare the computed F-Statistic with the critical value from F-distribution with dfB and dfW degree of freedom. If the computed F stat is greater than the critical value, then 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
In any statistical hypothesis test, the null hypothesis serves as a starting point and represents a statement of no effect or no difference. Here, the null hypothesis (\(H_0\)) posits that there is no difference in the mean highest number of drinks consumed across the seven different groups of drinkers. This means all group means are presumed equal.
When conducting an ANOVA test, the null hypothesis is crucial because it provides a baseline for comparison. The test essentially seeks to determine whether this hypothesis can be rejected based on the evidence provided by the data. If the null hypothesis holds true, any observed variation between group means is due to random sampling errors rather than a significant effect. Rejecting this hypothesis implies that at least one group has a different mean, indicating a potential difference in drinking patterns among the groups.
Sum of Squares
Sum of Squares is a key concept in ANOVA that helps in understanding the variability in data. There are two main types involved: **Sum of Squares Within (SSW)** and **Sum of Squares Between (SSB)**.
  • **Sum of Squares Within (SSW):** This measures the variation of observations within each group. It calculates how individual data points deviate from their group鈥檚 mean, indicating how consistently the groups' data are clustered around their mean. Mathematically, it's calculated using the formula: \[SSW = \sum_i^n s_i^2 * (n_i - 1)\] where \(s_i\) is the standard deviation of a group and \(n_i\) is the number of observations in that group.

  • **Sum of Squares Between (SSB):** This measures the variation between the group means themselves. It reflects how much the means differ from the overall mean of the data. It鈥檚 computed as: \[SSB = \sum_i^n n_i * (x_i - x_{total})^2\] where \(x_i\) is the mean of a group, and \(x_{total}\) is the total mean across all observations and groups.
Both sums of squares play a vital role in calculating the ANOVA test statistic and ultimately in determining whether the null hypothesis can be rejected.
F-Statistic
The F-Statistic is a crucial element in ANOVA testing. It is used to compare the variances and helps determine whether the differences between group means are statistically significant. The F-Statistic is computed using the ratio of the Mean Square Between (MSB) to the Mean Square Within (MSW).
The formula is:\[F = \frac{SSB / dfB}{SSW / dfW}\]Here:
  • **SSB** and **SSW** are the Sum of Squares Between and Within, respectively.
  • **dfB** (degrees of freedom between) is calculated as \( n - 1 \), where \( n \) is the number of groups.
  • **dfW** (degrees of freedom within) is found by \( \sum_i^n (n_i - 1) \).
A larger F-Statistic suggests that the variance between the group means is much larger than the variance within each group, pointing to a greater likelihood that the group means are indeed different, allowing for rejection of the null hypothesis. Conversely, a smaller F-Statistic implies that any observed differences might just be due to chance.
Degrees of Freedom
Degrees of Freedom are a key concept in statistics, particularly in determining the validity of various statistical operations. In the context of ANOVA, degrees of freedom are used to account for the variability in your data and are essential in calculating the F-Statistic. There are two types of degrees of freedom involved here: **df Between (dfB)** and **df Within (dfW)**.
  • **Degrees of Freedom Between (dfB):** This is calculated as \( n - 1 \), where \( n \) is the number of groups being compared. It represents the number of independent variations among the group means.

  • **Degrees of Freedom Within (dfW):** This is computed as \( \sum_i^n (n_i - 1) \). It accounts for the variability within the groups, focusing on individual data points relative to their group mean.
Understanding degrees of freedom is crucial because they shape the F-distribution against which the computed F-Statistic is compared. Higher degrees of freedom generally lead to a more reliable estimate of population parameters, which is why a precise calculation here is key for valid statistical inference in ANOVA.

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Most popular questions from this chapter

Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data reported in Consumer Reports.) $$\begin{array}{llllll} \text { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \\ \text { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \\ \text { Blue Bonnet } & 13.5 & 13.4 & 14.1 & 14.3 & \\ \text { Chiffon } & 13.2 & 12.7 & 12.6 & 13.9 & \\ \text { Mazola } & 16.8 & 17.2 & 16.4 & 17.3 & 18.0 \\ \text { Fleischmann's } & 18.1 & 17.2 & 18.7 & 18.4 & \end{array}$$ a. Test for differences among the true average PAPUFA percentages for the different brands. Use \(\alpha=.05\). b. Use the T-K procedure to compute \(95 \%\) simultaneous confidence intervals for all differences between means and interpret the resulting intervals.

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?

Give as much information as you can about the \(P\) -value of the single-factor ANOVA \(F\) test in each of the following situations. a. \(k=5, n_{1}=n_{2} \equiv n_{3}=n_{4}=n_{5}=4, F=5.37\) b. \(k=5, n_{1}=n_{2}=n_{3}=5, n_{4}=n_{5}=4, F=2.83\) c. \(k=3, n_{1}=4, n_{2}=5, n_{3}=6, F=5.02\) d. \(k=3, n_{1}=n_{2}=4, n_{3}=6, F=15.90\) e. \(k=4, n_{1}=n_{2}=15, n_{3}=12, n_{4}=10, F=1.75\)

Employees of a certain state university system can choose from among four different health plans. Each plan differs somewhat from the others in terms of hospitalization coverage. Four samples of recently hospitalized individuals were selected, each sample consisting of people covered by a different health plan. The length of the hospital stay (number of days) was determined for each individual selected. a. What hypotheses would you test to decide whether average length of stay was related to health plan? (Note: Carefully define the population characteristics of interest.) b. If each sample consisted of eight individuals and the value of the ANOVA \(F\) statistic was \(F=4.37\), what conclusion would be appropriate for a test with \(\alpha=.01\) ? c. Answer the question posed in Part (b) if the \(F\) value given there resulted from sample sizes \(n_{1}=9, n_{2}=8\), \(n_{3}=7\), and \(n_{4}=8\).

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.
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