91Ó°ÊÓ

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)

Step by step solution

01

- State the Hypotheses

First, state the null and alternative hypotheses. Null Hypothesis (H_0): The mean pH levels of rainwater are the same for Alaska, Florida, and Texas.Alternative Hypothesis (H_a): At least one mean pH level is different among the three states.

02

- Verify ANOVA Requirements

Verify that the data meets the requirements for performing a one-way ANOVA: 1. The samples are independent. 2. The data comes from normally distributed populations (this is given). 3. The variances of the populations are approximately equal. To check for equal variances, use Levene’s Test. This can be done using statistical software.

03

- Perform the One-Way ANOVA Test

Use the following data: Alaska: [5.41, 5.39, 4.90, 5.14, 4.80, 5.24]Florida: [4.87, 5.18, 4.52, 5.12, 4.89, 5.06]Texas: [5.46, 5.89, 5.57, 5.15, 5.45, 5.30]Calculate the means and variances of each group. Perform the one-way ANOVA test using statistical software or a calculator using these steps:1. Calculate the overall mean 2. Calculate the sum of squares between groups (SSB) 3. Calculate the sum of squares within groups (SSW) Then apply the F-test: F = (SSB / (k-1)) / (SSW / (n-k)) where k is the number of groups (3 in this case) and n is the total number of observations (18 in this case). Compare the F-value against the critical value from the F-distribution table with df1 = k-1 and df2 = n-k at α=0.05.

04

- Interpret the Results

If the computed F-value is greater than the critical F-value from the F-distribution table, reject the null hypothesis (H_0); otherwise, do not reject it.

05

- Draw Boxplots

Draw boxplots for visual comparison of the pH levels in rainwater for Alaska, Florida, and Texas. The boxplot should show the median, quartiles, and any potential outliers in the data for each state. This will support the conclusion from the ANOVA test.

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Key Concepts

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

Null Hypothesis

In statistical testing, we often begin by assuming there is no effect or no difference. This assumption is called the null hypothesis, which we denote as H0. For the environmentalist's study, the null hypothesis is that the mean acidity (pH level) of rainwater is the same across three states: Alaska, Florida, and Texas.
The null hypothesis helps us frame our investigation. By testing this assumption, we can use statistical methods to either reject it or fail to reject it based on the data sampled. If sufficient evidence from the samples is gathered, the null hypothesis can be rejected, indicating an effect or difference does exist.

Alternative Hypothesis

Opposite to the null hypothesis is the alternative hypothesis, denoted as Ha. It represents what we aim to provide evidence for through our analysis. In this context, the alternative hypothesis claims that at least one state among Alaska, Florida, and Texas has a mean pH level that is different.
Thus, while the null hypothesis assumes all means are the same, the alternative hypothesis suggests a significant discrepancy among at least one pair of the states. The objective of hypothesis testing will be to determine if there is enough statistical evidence to support this alternative view.

One-Way ANOVA

The one-way ANOVA (Analysis of Variance) is a technique used when comparing the means of three or more independent groups. The goal is to determine if there is a statistically significant difference between the group means. In our environmental study, we use one-way ANOVA to compare the acidity levels of rainwater among Alaska, Florida, and Texas.
One-way ANOVA operates by partitioning the overall variance into components associated with the differences between group means and within group variances. Essentially, it helps in testing the equality of means across the groups by analyzing varied sources of data variance.
To check if the assumptions of ANOVA are met, requirements such as normality of data distribution and equal variances across groups (homogeneity of variances) are verified.

Levene’s Test

Before running the ANOVA, it’s critical to check if the variances across the groups are approximately equal. This is where Levene’s Test comes into play. This statistical test is used to evaluate if the variances across several groups are homogeneous.
To perform Levene’s Test, the absolute deviations from the group means are computed, and then the hypothesis about equal variances is assessed using an F-test on these deviations. If Levene’s Test shows that the variances are not significantly different, the assumption of homogeneity of variances holds true, and ANOVA results will be valid.

F-test

Within the realm of ANOVA, the F-test is employed to compare the variance between groups to the variance within groups. For this exercise, the F-test will help determine if the differences in pH levels across Alaska, Florida, and Texas are statistically significant.
Here's how the F-test works in ANOVA:

• Compute the sum of squares between (SSB) groups – a measure of variance among the sample means.
• Compute the sum of squares within (SSW) groups – a measure of variance within each sample.
• Calculate the F-value: \[\text{F} = \frac{\frac{SSB}{k-1}}{\frac{SSW}{n-k}}\] where k is the number of groups and n is the total number of observations.
• The F-value is then compared with a critical value from the F-distribution table. If the calculated F-value is greater than the table’s critical F-value at a given significance level (like 0.05), we reject the null hypothesis.

The F-test is integral to ANOVA, guiding the determination of whether observed data supports the claim of differences among population means.