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Got Milk? Researchers Sharon Peterson and Madeleine Sigman-Grant wanted to compare the overall nutrient intake of American children (ages 2 to 19 ) who exclusively use skim milk instead of \(1 \%, 2 \%,\) or whole milk. The researchers combined children who consumed \(1 \%\) or \(2 \%\) milk into a "mixed milk" category. The following data represent the daily calcium intake (in \(\mathrm{mg}\) ) for a random sample of eight children in each category and are based on the results presented in their article "Impact of Adopting Lower-Fat Food Choices on Nutrient Intake of American Children," Pediatrics, Vol. \(100,\) No. \(3 .\) $$ \begin{array}{ccc} \text { Skim Milk } & \text { Mixed Milk } & \text { Whole Milk } \\ \hline 916 & 1024 & 870 \\ \hline 886 & 1013 & 874 \\ \hline 854 & 1065 & 881 \\ \hline 856 & 1002 & 836 \\ \hline 857 & 1006 & 879 \\ \hline 853 & 991 & 938 \\ \hline 865 & 1015 & 841 \\ \hline 904 & 1035 & 818 \\ \hline \end{array} $$ (a) Is there sufficient evidence to support the belief that at least one of the means is different from the others at the \(\alpha=0.05\) level of significance? Note: The requirements for a one-way ANOVA are satisfied. (b) If the null hypothesis is rejected in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (c) Draw boxplots of the three categories to support the analytic results obtained in parts (a) and (b).

Short Answer

Expert verified
Reject the null hypothesis if F > F_critical. Perform Tukey's test to identify which means differ.

Step by step solution

01

- State the hypotheses

Define the null and alternative hypotheses for the one-way ANOVA test. The null hypothesis ( H_0 ) is that all the means are equal, and the alternative hypothesis ( H_a ) is that at least one mean is different. H_0: μ_1 = μ_2 = μ_3 H_a: At least one μ is different
02

- Calculate the means and variances

Compute the sample means and variances for each of the three categories: Skim Milk, Mixed Milk, and Whole Milk.Skim Milk: mean_1 = (916 + 886 + 854 + 856 + 857 + 853 + 865 + 904) / 8 = 861.375 Mixed Milk: mean_2 = (1024 + 1013 + 1065 + 1002 + 1006 + 991 + 1015 + 1035) / 8 = 1006.375 Whole Milk: mean_3 = (870 + 874 + 881 + 836 + 879 + 938 + 841 + 818) / 8 = 867.125
03

- Calculate the ANOVA test statistic

Using the sample means and variances, calculate the ANOVA test statistic (F) and the critical value at α=0.05. Since the detailed calculation can be lengthy, summarize the key formulas: SSB (Sum of squares between groups) SSW (Sum of squares within groups) MSB (Mean squares between groups) = SSB / (k-1) MSW (Mean squares within groups) = SSW / (N-k) F = MSB/MSW
04

- Compare F to the critical value

Find the critical value for F at α=0.05 and degree of freedom (k-1, N-k). Since the calculation involves F-distribution tables or tools, the important part is to compare the calculated F with the critical value from the table. If F is greater than the critical value, reject the null hypothesis (H_0).
05

- Conduct Tukey's test

If null hypothesis is rejected, apply Tukey's test for multiple comparisons to find out which specific group means are significantly different from each other. This involves calculating the differences between each pair of means and comparing them to the Tukey's critical Q value.
06

- Draw boxplots

Draw boxplots for the three categories (Skim Milk, Mixed Milk, Whole Milk) to visually inspect the differences in distribution and support the analytic results from the ANOVA and Tukey's test.

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Key Concepts

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

ANOVA Test
The Analysis of Variance (ANOVA) test is a statistical method used to compare the means of three or more groups to determine if there is a significant difference among them. We use a one-way ANOVA when we want to test for differences among means in a single factor (e.g., types of milk).
In our exercise, the null hypothesis (\(H_0\)) states that all group means are equal, while the alternative hypothesis (\(H_a\)) suggests at least one group mean differs.
Steps involved in conducting a one-way ANOVA include:
  • Calculating the mean and variance for each group
  • Computing the ANOVA test statistic (F)
  • Comparing the F value with the critical value from the F-distribution table
  • If F is greater than the critical value, reject the null hypothesis

Remember to check assumptions: normality, homogeneity of variances, and independence of observations.
Calcium Intake
Calcium intake plays a crucial role in the health and development of children. It's essential for bone growth and maintenance. Different milk types provide varying levels of calcium. Hence, the researchers in the exercise aimed to compare calcium intake among children consuming different types of milk: skim, mixed, and whole milk.
By analyzing calcium intake levels, you can understand whether certain milk types contribute significantly more calcium to children's diets. This information is vital for dietary recommendations and ensuring children meet their nutritional requirements.
Statistical Hypothesis Testing
Statistical hypothesis testing is a method of making decisions using data. It involves proposing a hypothesis and then using statistical tools to determine the likelihood that the hypothesis is true.
In our exercise:
  • The null hypothesis (\(H_0\)) suggests no difference in calcium intake among the groups.
  • The alternative hypothesis (\(H_a\)) states that at least one group has a different mean calcium intake.
One-way ANOVA helps in testing these hypotheses by analyzing the variance within and between the groups.
Steps include:
  • Stating null and alternative hypotheses
  • Choosing a significance level (α=0.05)
  • Calculating the test statistic
  • Comparing it against a critical value
If the test statistic exceeds the critical value, the null hypothesis is rejected, indicating significant differences.
Tukey's Test
Tukey's test, or Tukey's Honest Significant Difference (HSD) test, is post-hoc analysis used after an ANOVA test to find out which specific group means (pairs) are significantly different.
Steps include:
  • Calculating the absolute difference between each pair of group means
  • Comparing these differences to the Tukey's critical Q value
If the observed mean difference exceeds the Tukey critical value, it indicates significant differences between those pairs.
In our exercise, Tukey's test follows after rejecting the null hypothesis in ANOVA, enabling identification of specific groups (skim, mixed, or whole milk) that have different mean calcium intakes.
Boxplot
Boxplots are visual tools for graphically depicting groups of numerical data through their quartiles. In this exercise, they illustrate the distribution of calcium intake across the milk types.
Each boxplot shows:
  • The median
  • Upper and lower quartiles
  • Possible outliers
  • Range

By drawing boxplots for skim, mixed, and whole milk categories, you can visually compare their spreads and medians. This visual inspection supports ANOVA results by showcasing similarities or differences in calcium intake distributions across different milk types.

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Most popular questions from this chapter

The Insurance Institute for Highway Safety conducts experiments in which cars are crashed into a fixed barrier at \(40 \mathrm{mph}\). The barrier's deformable face is made of aluminum honeycomb, which makes the forces in the test similar to those involved in a frontal offset crash between two vehicles of the same weight, each going just less than \(40 \mathrm{mph}\). Suppose you want to know if the mean head injury resulting from this offset crash is the same for large family cars, passenger vans, and midsize utility vehicles. The researcher wants to determine if the means for head injury for each class of vehicle are different. The following data were collected from the institute's study. $$\begin{array}{lc}\text { Large Family Cars } & \text { Head Injury (hic) } \\\ \hline \text { Hyundai XG300 } & 264 \\\\\hline \text { Ford Taurus } & 134 \\\\\hline \text { Buick LeSabre } & 409 \\\\\hline \text { Chevrolet Impala } & 530 \\\\\hline \text { Chrysler 300 } & 149 \\\\\hline \text { Pontiac Grand Prix } & 627 \\\\\hline \text { Toyota Avalon } & 166 \\\\\text { Passenger Vans } & \text { Head Injury (hic) } \\\\\hline \text { Toyota Sienna } & 148 \\\\\hline \text { Honda Odyssey } & 238 \\\\\hline \text { Ford Freestar } & 340 \\\\\hline \text { Mazda MPV } & 693 \\\\\hline \text { Chevrolet Uplander } & 550 \\\\\hline \text { Nissan Quest } & 470 \\\\\hline \text { Kia Sedona } & 322\end{array}$$ $$\begin{array}{lc}\hline \text { Midsize Utility Vehicles } & \text { Head Injury (hic) } \\\\\hline \text { Honda Pilot } & 225 \\\\\hline \text { Toyota 4Runner } & 216 \\\\\hline \text { Mitsubishi Endeavor } & 186 \\\\\hline \text { Nissan Murano } & 307 \\\\\hline \text { Ford Explorer } & 353 \\\\\hline \text { Kia Sorento } & 552 \\\\\hline \text { Chevy Trailblazer } & 397 \\\\\hline\end{array}$$ (a) State the null and alternative hypotheses. (b) Verify that the requirements to use the one-way ANOVA procedure are satisfied. Normal probability plots indicate that the sample data come from normal populations. (c) Test the hypothesis that the mean head injury for each vehicle type is the same at the \(\alpha=0.01\) level of significance. (d) Draw boxplots of the three vehicle types to support the analytic results obtained in part(c).

The following data are taken from three different populations known to be normally distributed, with equal population variances based on independent simple random samples. $$ \begin{array}{ccc} \text { Sample 1 } & \text { Sample 2 } & \text { Sample 3 } \\ \hline 35.4 & 42.0 & 43.3 \\ \hline 35.0 & 39.4 & 48.6 \\ \hline 39.2 & 33.4 & 42.0 \\ \hline 44.8 & 35.1 & 53.9 \\ \hline 36.9 & 32.4 & 46.8 \\ \hline 28.9 & 22.0 & 51.7 \\ \hline \end{array} $$ (a) Test the hypothesis that each sample comes from a population with the same mean at the \(\alpha=0.05\) level of significance. That is, test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) (b) If you rejected the null hypothesis in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (c) Draw boxplots of each set of sample data to support your results from parts (a) and (b).

Given the following ANOVA output, answer the questions that follow: $$ \begin{aligned} &\text { Analysis of Variance for Response }\\\ &\begin{array}{lrrrrr} \text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Block } & 6 & 1712.37 & 285.39 & 134.20 & 0.000 \\ \text { Treatment } & 3 & 2.27 & 0.76 & 0.36 & 0.786 \\ \text { Error } & 18 & 38.28 & 2.13 & & \\ \text { Total } & 27 & 1752.91 & & & \end{array} \end{aligned} $$ (a) The researcher wants to test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\) against \(H_{1}:\) at least one of the means is different. Based on the ANOVA table, what should the researcher conclude? (b) What is the mean square due to error? (c) Explain why it is not necessary to use Tukey's test on these data.

Given the following ANOVA output, answer the questions that follow: $$\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\\\\text { Block } & 4 & 768.27 & 192.067 & 10.96 & 0.002 \\\\\text { Treatment } & 2 & 278.53 & 139.267 & 7.95 & 0.013 \\\\\text { Error } & 8 & 140.13 & 17.517 & & \\\\\text { Total } & 14 & 1186.93 & & &\end{array}$$ (a) The researcher wants to test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) against \(H_{1}:\) at least one of the means is different. Based on the ANOVA table, what should the researcher conclude? (b) What is the mean square due to error? (c) The following output represents the results of Tukey's test. What should the researcher conclude? Tukey simultaneous Tests Response Variable Response All Pairwise Comparisons among Levels of Treatment Treatment \(=1\) subtracted from: Treat- Difference \(\quad\) SB of \(\quad\) Adj usted me of Means Difference \(T\) -Value P-Value 2 -1.000 \(2.647 \quad-0.3778\) 0.9251 3 8.600 2.647 3.2489 0.0282 Treatment \(=2\) subtracted from Treat- Difference \(\quad\) SE of \(\quad\) Adjusted ment \(\begin{array}{llll}\text { of Means Difference } & \text { T-Value } & \text { P-Value }\end{array}\) \(\begin{array}{lllll}3 & 9.600 & 2.647 & 3.627 & 0.0165\end{array}\)

The comparisonwise error rate, denoted \(\alpha_{c}\), is the probability of making a Type I error when comparing two means. It is related to the familywise error rate, \(\alpha\), through the formula \(1-\alpha=\left(1-\alpha_{c}\right)^{k},\) where \(k\) is the number of means being compared. (a) If the familywise error rate is \(\alpha=0.05\) and \(k=3\) means are being compared, what is the comparisonwise error rate? (b) If the familywise error rate is \(\alpha=0.05\) and \(k=5\) means are being compared, what is the comparisonwise error rate? (c) Based on the results of parts (a) and (b), what happens to the comparisonwise error rate as the number of means compared increases?

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