12.3 MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table 12.8 showsvarious colony counts from different patients who may or may not have MRSA. The data from the table is plotted inFigure 12.5.Conc = 0.6 Conc = 0.816938296698Table 12.8Plot of the data for the different concentrations:Tryptone concentrations1.41.21-0.80.6 .50Conc = 1.022147120Conc = 1.230199148100Colony counts150Conc = 1.427168132200Figure 12.5Test whether the mean number of colonies are the same or are different. Construct the ANOVA table, find the p-value,and state your conclusion. Use a 5% significance level.

The objective of this question is to test whether the mean number of colonies are the same or different for different concentrations using Analysis of Variance (ANOVA).First, we need to calculate the mean of each group. The means are as follows: For concentration 0.6, the mean is (16+93+82)/3 = 63.67 For concentration 0.8, the mean is (9+66+98)/3 = 57.67 For concentration 1.0, the mean is (22+147+120)/3 = 96.33 For concentration 1.2, the mean is (30+199+148)/3 = 125.67 For concentration 1.4, the mean is (27+168+132)/3 = 109Next, we calculate the overall mean, which is the sum of all observations divided by the total number of observations. The overall mean is (63.67+57.67+96.33+125.67+109)/5 = 90.47Now, we calculate the sum of squares between groups (SSB), which is the sum of the squared differences between each group mean and the overall mean. SSB = 3*((63.67-90.47)^2 + (57.67-90.47)^2 + (96.33-90.47)^2 + (125.67-90.47)^2 + (109-90.47)^2) = 3*339.64 = 1018.92Next, we calculate the sum of squares within groups (SSW), which is the sum of the squared differences between each observation and its group mean. SSW = (16-63.67)^2 + (93-63.67)^2 + (82-63.67)^2 + (9-57.67)^2 + (66-57.67)^2 + (98-57.67)^2 + (22-96.33)^2 + (147-96.33)^2 + (120-96.33)^2 + (30-125.67)^2 + (199-125.67)^2 + (148-125.67)^2 + (27-109)^2 + (168-109)^2 + (132-109)^2 = 10183.33The degrees of freedom between groups (dfB) is the number of groups minus 1, which is 5-1 = 4. The degrees of freedom within groups (dfW) is the total number of observations minus the number of groups, which is 15-5 = 10.The mean square between groups (MSB) is SSB/dfB = 1018.92/4 = 254.73. The mean square within groups (MSW) is SSW/dfW = 10183.33/10 = 1018.33The F statistic is MSB/MSW = 254.73/1018.33 = 0.25The p-value is the probability of observing a value as extreme as the F statistic under the null hypothesis. We can find this value using a F-distribution table or a statistical software. For this F statistic and degrees of freedom, the p-value is approximately 0.91Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the mean number of colonies are different for different concentrations.The ANOVA test was conducted to determine if the mean number of colonies are different for different concentrations. The p-value of 0.91 is greater than the significance level of 0.05, so we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the mean number of colonies are different for different concentrations.