Use the data below, showing a summary of highway gas mileage for several observations, to decide if the average highway gas mile

Hello! To compare average highway gas mileage across three vehicle types (midsize cars, SUVs, and pickup trucks) at the 1% significance level, you need to conduct an ANOVA test. The relevant data are as follows: n | Mean | Std. Dev. Midsize | 31 | 25.8 | 2.56 SUV’s | 31 | 22.68 | 3.67 Pickups | 14 | 21.29 | 2.76 Defining variables: X₁: highway mileage of a midsize car X₂: highway mileage of an SUV X₃: highway mileage of a pickup truck. Assuming the variables display a normal distribution and independence, we can set our hypotheses: H₀: μ₁ = μ₂ = μ₃ H₁: At least one of the population means differs. Significance level α is 0.01. The test statistic is computed as shown in the attached ANOVA table. For treatments, the degrees of freedom between treatments are k-1, with k being the total number of treatments. Next, the sum of squares for treatments can be calculated. SSTr = ∑ni(Ÿi - Ÿ..)², where ŷi represents the sample mean for each sample i and ŷ.. indicates the grand mean. Calculate the grand mean by summing the group means divided by the total number of groups: ŷ..= (Ÿ₁ + ŷ₂ + ŷ₃)/3 = (25.8+22.68+21.29)/3 = 23.256 ≈ 23.26. Sum of squares: = (Ÿ₁ - ŷ..)² + (Ÿ₂ - ŷ..)² + (Ÿ₃ - ŷ..)² = (25.8-23.26)² + (22.68-23.26)² + (21.29-23.26)² = 10.6689. For errors, the degrees of freedom for errors are: Mean square equals the variance estimation of errors; this can be computed using a predefined formula. The rejection region for this test is always one-tailed to the right, meaning you reject the null hypothesis for significantly large values of the statistic: If the calculated statistic is ≥ 4.07, reject the null hypothesis; otherwise, do not reject it. In this scenario, since the statistic is less than the critical value, the conclusion is to not reject the null hypothesis. At 1% significance, we conclude that the average highway mileage remains consistent across the three vehicle types.