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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).

Short Answer

Expert verified
Perform ANOVA to test if sample means are equal. If rejected, use Tukey's test to find pair differences. Boxplots support results.

Step by step solution

01

- State the Hypotheses

We are testing the null hypothesis that all three samples come from populations with the same mean. Formally, the hypotheses are: \[ H_{0}: \, \mu_{1} = \mu_{2} = \mu_{3} \]\[ H_{a}: \, \text{At least one population mean is different} \]
02

- Calculate the ANOVA

To test the hypothesis, perform a one-way ANOVA (Analysis of Variance). This requires calculation of the between-group and within-group variances. ANOVA will help determine whether the differences in sample means are statistically significant.
03

- Summarize Data and ANOVA Table

Calculate the sum of squares between groups (SSB), sum of squares within groups (SSW), degrees of freedom, and mean squares (MSB and MSW). Then, calculate the F-ratio:\[ F = \frac{MSB}{MSW} \]Compare the calculated F-value with the critical F-value from the F-distribution table at \( \alpha=0.05 \) and degrees of freedom corresponding to the samples.
04

- Decision Rule

If the calculated F-value is greater than the critical F-value, reject the null hypothesis \( H_{0} \). Otherwise, fail to reject \( H_{0} \).
05

Step 5a - Conclusion for Part (a)

Determine whether the null hypothesis is rejected or not, based on the F-test results from Step 4. If rejected, move to Tukey's test to find which pairwise means differ.
06

Step 5b - Perform Tukey鈥檚 Test (if needed)

If the null hypothesis was rejected, use Tukey's Honestly Significant Difference (HSD) test to determine which sample means are significantly different. Calculate the HSD value and compare it to the differences between sample means.
07

- Boxplots

Draw boxplots for each sample. The boxplots visually show the spread and center of each sample data set, which can support the ANOVA and Tukey's test results.

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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 foundational concept in statistics. It represents the default or standard position that there is no difference or effect.
In the context of ANOVA, the null hypothesis (ull hypothesisull hypothesisull hypothesis ull hypothesisull hypothesis) is that all population means are equal.
  • Formulated as: ull hypothesis: 渭鈧 = 渭鈧 = 渭鈧
  • Indicates no significant difference between the sample means.
  • For our exercise, we will either reject or not reject this hypothesis based on the ANOVA test.

The alternative hypothesis (ull hypothesisull hypothesis) states that ull hypothesis one population mean is different.
To make a decision, we usually compare our test statistic to a critical value.
  • If our calculated value exceeds the critical value, we reject ull hypothesisull hypothesis.
  • Otherwise, we do not reject ull hypothesisull hypothesis.
one-way ANOVA
A one-way ANOVA (Analysis of Variance) is used to determine if there are statistically significant differences between the means of three or more independent groups.

Here鈥檚 a simplified process for the ANOVA test:
  • Calculate the means of each group.
  • Measure the variation between the groups (Sum of Squares Between, SSB).
  • Measure the variation within the groups (Sum of Squares Within, SSW).
  • Divide each sum of squares by their respective degrees of freedom to get Mean Squares (MSB and MSW).
  • Calculate the F-ratio: F = MSB / MSW
By comparing the F-ratio to a critical value from the F-distribution table, you can decide whether to reject the null hypothesis.

If ull hypothesis: ull hypothesis鈮0 and the calculated F-value is higher than the critical value, you reject the null hypothesis, indicating at least one group mean is different.
Tukey's test
When ANOVA shows significant differences, Tukey's Honestly Significant Difference (HSD) test helps determine which specific groups differ.

Steps for Tukey's test:
  • Calculate the mean difference for each pair of groups.
  • Compute the HSD value, which accounts for the number of comparisons.
  • Compare each mean difference to the HSD value.
If a mean difference is greater than the HSD value, it is considered significant.

This test keeps the overall error rate (ull hypothesisull hypothesis) at the specified level (usually 0.05), minimizing the chance of false positives when making multiple comparisons.
boxplots
Boxplots are graphical representations that summarize data using quartiles and medians. They provide a visual overview of data distribution.

Each boxplot shows:
  • The median (middle line of the box).
  • The interquartile range (IQR) 鈥 the box contains the middle 50% of the data.
  • Whiskers 鈥 lines extending from the box to the smallest and largest values within 1.5 * IQR from the quartiles.
  • Outliers 鈥 points outside the whiskers.
Boxplots help in comparing distributions across groups. In our exercise, drawing boxplots for each sample set aids in visually confirming the ANOVA and Tukey's test results.
  • You can see if the medians and spreads of sample data align with statistical findings.
  • Notable differences in the boxplots often indicate significant differences found by the tests.

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Most popular questions from this chapter

An environmentalist wanted to determine if the mean acidity of rain differed among Alaska, Florida, and Texas. He randomly selected six rain dates at each of the three locations and obtained the following data:$$\begin{array}{ccc}\text { Alaska } & \text { Florida } & \text { Texas } \\\\\hline 5.41 & 4.87 & 5.46 \\\\\hline 5.39 & 5.18 & 5.89 \\\\\hline 4.90 & 4.52 & 5.57 \\\\\hline 5.14 & 5.12 & 5.15 \\\\\hline 4.80 & 4.89 & 5.45 \\\\\hline 5.24 & 5.06 &5.30\end{array}$$ (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 a normal population. (c) Test the hypothesis that the mean pHs in the rainwater are the same at the \(\alpha=0.05\) level of significance. (d) Draw boxplots of the \(\mathrm{pH}\) in rain for the three states to support the results obtained in part (c)

Suppose there is sufficient evidence to reject \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}\) using a one-way ANOVA. The mean square error from ANOVA is determined to be \(26.2 .\) The sample means are \(\bar{x}_{1}=42.6, \bar{x}_{2}=49.1, \bar{x}_{3}=46.8, \bar{x}_{4}=63.7,\) with \(n_{1}=n_{2}=n_{3}=n_{4}=6 .\) Use Tukey's test to determine which pairwise means are significantly different using a familywise error rate of \(\alpha=0.05 .\)

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).

Nutrition Researchers Sharon Peterson and Madeleine Sigman-Grant wanted to compare the overall nutrient intake of American children (ages 2 to 19 ) who exclusively use lean meats, mixed meats, or higher-fat meats. The data given on the next page represent the daily consumption of calcium (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. \(3 .\) $$ \begin{array}{ccc} \text { Lean Meats } & \text { Mixed Meats } & \text { Higher-Fat Meats } \\ \hline 844.2 & 897.7 & 843.4 \\ \hline 745.0 & 908.1 & 862.2 \\ \hline 773.1 & 948.8 & 790.5 \\ \hline 823.6 & 836.6 & 876.5 \\ \hline 812.0 & 871.6 & 790.8 \\ \hline 758.9 & 945.9 & 847.2 \\ \hline 810.7 & 859.4 & 772.0 \\ \hline 790.6 & 920.2 & 851.3 \\ \hline \end{array} $$ (a) The data was collected using a cohort observational study. Explain what this means. (b) What is the response variable in this study? (c) What are the null and alternative hypotheses in this study? (d) Test the null hypothesis that the mean calcium for each category is the same at the \(\alpha=0.05\) level of significance. Note: The requirements for a one-way ANOVA are satisfied. (e) If the null hypothesis is rejected in part (d), use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (f) Draw boxplots of the three treatment levels to support the analytic results obtained in parts (d) and (e). (g) Can any statements of causality between meat consumption and consumption of calcium be made based on the results of this study? Explain.

Explain what an interaction effect is. Why is it dangerous to analyze main effects if there is an interaction effect?

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