**ANOVA Analysis of Electrical Resistivity for Concrete Mixtures**An experiment was conducted to compare electrical resistivity for six different low-permeability concrete bridge deck mixtures. Each mixture was measured 26 times on concrete cylinders, 28 days post-casting. The table below represents an ANOVA (Analysis of Variance) based on data from an article. Complete the missing entries, rounding the f value to two decimal places.**ANOVA Table:**| Source | df | Sum of Squares | Mean Square | f ||----------|-----|----------------|-------------|--------|| Mixture | 5 | | | || Error | 150 | | 13.969 | || Total | 155 | 5664.475 | | |**Instructions:**Use the Distribution Calculators page in SALT to find critical values and/or p-values to answer parts of this question.**Hypothesis Testing at Level 0.05:**State the appropriate hypotheses. Let \( \mu_i \) represent the true mean electrical resistivity for the i-th mixture.- \( H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5 = \mu_6 \)- \( H_a \): at least two \( \mu_i \)'s are unequal**Select the Appropriate Hypothesis:**- \( H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4 = \mu_5 = \mu_6 \)- \( H_a \): at least two \( \mu_i \)'s are unequal(The correct option is selected with a checkmark.)**Test Statistic Calculation:**Round your answer to two decimal places for f = ...**P-Value Interpretation:**Discuss what can be concluded about the p-value for the test.**Explanation:**The ANOVA table analysis involves testing the equality of means for different groups. The mean square values are calculated for both the treatments (mixtures) and the error, and the f statistic is determined. The hypothesis test checks if there is a statistically significant difference among the means of the mixtures, where a significant f value suggests rejecting the null hypothesis.

Answer Given The number of mixtures = 6K = 6The number of observations, n = 26DF [Mixture] = K- 1 = 5DF [Error ] =K* [n-1] =150DF [Total] = DF [Mixture] + DF [Error ] = 150+5 = 155SS [Total] = 5664.475MS [Error] = 13.969DF [Error] = 150SS [Error] = MS[Error] * DF [Error] = 13.969 * 150 = 2095.35SS [Mixture] = SS [Total] - SS [Error] = 5664.475 - 2095.35 = 3569.125MS [Mixture] = SS [Mixture] / [DF [mixture]] = 3569.125 / 5 = 713.825The test statistics:F = [MS [Mixture]]/ [MS [Error]] = [713.825] / [13.969] = 51.10P-value:F.DIST. RT [51.10, 5, 150]= 0.000P<0.05Reject the H0,There is sufficient evidence to conclude the claim at 0.05.Final answer ANOVA table:Source DF SS MS F PMixture 5 3569.125 713.825 51.10 0.0000Error 150 2095.35 13.969Total 155 5664.475P<0.05reject the H0,Sufficient evidence