**8.6** A large laboratory has four types of devices used to determine the pH of soil samples. The laboratory wants to determine whether there are differences in the average readings given by these devices. The lab uses 24 soil samples having known pH in the study and randomly assigns six of the samples to each device. The soil samples are tested, and the response recorded for each sample is the difference between the pH reading of the device and the known pH of the soil. These values, along with summary statistics, are given in the following table.| Device | Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample 5 | Sample 6 ||--------|----------|----------|----------|----------|----------|----------|| A | -0.307 | -0.294 | 0.079 | 0.019 | -0.136 | -0.324 || B | -0.176 | 0.125 | -0.013 | 0.082 | 0.091 | 0.459 || C | 0.137 | -0.063 | 0.240 | -0.050 | 0.318 | 0.154 || D | -0.042 | 0.690 | 0.201 | 0.166 | 0.219 | 0.407 |**Analysis** Based on your intuition, is there evidence to indicate any difference among the mean differences in pH readings for the four devices? Run an analysis of variance to confirm or reject your conclusion. Use α = 0.05.

The given data is as follows Sample123456A-0.307-o.2940.0790.019-0.136-0.324B-0.1760.125-0.0130.0820.0910.459C0.137-0.0630.240-0.0500.3180.154D-0.0420.6900.2010.1660.2190.407Null hypothesis H0:The four device means are equal.There is no significant difference among the mean differences in PH readings for the four devices.Alternative hypothesis H1:At least one mean is not equalThere is a significant difference among the mean differences in PH readings for the four devices.Calculate the following values for performing one-way ANOVAFrom the given information number of treatments k=4 (Number of devices) Total number of observations =24Degrees of freedom:Degrees of freedom for between devices=Number of devices -1=4-1=3Degrees of freedom for within devices=Total number of observations -number of devices =N-k=24-4=20Mean sum of squares:Mean sum of squares for between devices = sum of squares for between devices/Degrees of freedom for between devicesMean sum of squares for within devices =sum of squares for within devices/Degrees of freedom for within devicesF-Ratio:F-Ratio =Mean sum of squares for between devices /Mean sum of squares for within devicesConstructing the following table for finding the required values of ANOVA 123456Ti.Ti.2Ti.2/niA-0.307-0.2940.0790.019-0.136-0.324-0.9630.9273690.1545620.310759B-0.1760.125-0.0130.0820.0910.4590.5680.3226240.0537710.272456C0.137-0.0630.24-0.050.3180.1540.7360.5416960.0902830.207678D-0.0420.690.2010.1660.2190.4071.6412.6928810.4488140.7594311.9820.747428Total=1.550324From the above tableG=1.982RSS=1.550324computed values are arranged in the following tableSource of variationsum of squaresdegrees of freedomMean sum of squaresF-RatioBetween devices0.583748k-1=4-1=30.583748/3=0.194583F=0.194583/0.04014=4.8476~F(320)Within devices0.802896N-k=24-4=200.802896/20=0.04014Total1.386644N-1=24-1=23From the above ANOVA tableFcal =4.85compute the critical value of F at (3,20) degrees of freedom at a 5% level of significance from Excel.The Excel function for the critical value of F is =F.INV.RT(0.05,3,20) =3.098391Comparing the critical value and calculated value of FFcal>Fcri May reject the null hypothesis.Accept alternative hypothesis.We conclude that there is evidence to indicate any differences among the mean differences in PH reading for the four devices.There is evidence to indicate any differences among the mean differences in PH reading for the four devices.