In Exercises 21 and 22, (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal.
[APPLET] The table shows the monthly electric bills (in dollars) for a sample of households from four regions of the United States. At α=0.10, can you conclude that the mean monthly electric bill is different in at least one of the regions? (Adapted from U.S. Energy Information Administration)

Verified step by step guidance

1. Step 1: Identify the problem type and hypotheses. This is a one-way ANOVA test because we are comparing the means of more than two independent groups (Northeast, Midwest, South, West). The null hypothesis \(H_0\) is that all group means are equal: \(\mu_{Northeast} = \mu_{Midwest} = \mu_{South} = \mu_{West}\). The alternative hypothesis \(H_a\) is that at least one group mean is different.
2. Step 2: Calculate the sample means and variances for each region using the data provided. This involves computing the mean \(\bar{x}_i\) and variance \(s_i^2\) for each group \(i\) (Northeast, Midwest, South, West).
3. Step 3: Compute the overall mean \(\bar{x}\) by combining all data points from the four groups. Then calculate the Between-Group Sum of Squares (SSB) and Within-Group Sum of Squares (SSW) using the formulas: \(SSB = \sum_{i=1}^k n_i (\bar{x}_i - \bar{x})^2\) \(SSW = \sum_{i=1}^k (n_i - 1) s_i^2\) where \(k\) is the number of groups and \(n_i\) is the sample size of group \(i\).
4. Step 4: Calculate the Mean Squares: Mean Square Between (MSB) and Mean Square Within (MSW) by dividing the sums of squares by their respective degrees of freedom: \(MSB = \frac{SSB}{k-1}\) \(MSW = \frac{SSW}{N-k}\) where \(N\) is the total number of observations.
5. Step 5: Compute the F-statistic as \(F = \frac{MSB}{MSW}\). Compare this F-value to the critical value from the F-distribution table at significance level \(\alpha = 0.10\) with degrees of freedom \(df_1 = k-1\) and \(df_2 = N-k\). If \(F\) is greater than the critical value, reject the null hypothesis and conclude that there is sufficient evidence to say the mean monthly electric bill differs in at least one region. Otherwise, fail to reject the null hypothesis.

Key Concepts

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to compare the means of three or more groups to determine if at least one group mean is significantly different from the others. It tests the null hypothesis that all group means are equal against the alternative that at least one differs. This method is appropriate here because we are comparing electric bills across four regions.

Assumptions of ANOVA

ANOVA relies on key assumptions: samples must be random and independent, populations should be normally distributed, and population variances must be equal (homogeneity of variance). These assumptions ensure the validity of the test results and affect the interpretation of the p-value and conclusions.

Significance Level (α) and Hypothesis Testing

The significance level α (here 0.10) is the threshold for deciding whether to reject the null hypothesis. If the p-value from ANOVA is less than α, we conclude there is sufficient evidence that at least one region's mean electric bill differs. This decision must be interpreted in the context of the original claim about differences among regions.