Question

circumstances, the four laboratories should show no signifcant difference. The measurements from four laboratories are given in Table 1. Test hypothesis 1 at au5. To that end, enter measurements into SPSS such that you have two columns: the first column will con- tain the laboratory ID and the second column will contain the corresponding measurements as illustrated by Table 2. Table 2: laboratory measurement 1 001434 0.10511 1 0.27031 2 0.00290 2 0.00739 2 0.35320 Choose AnalyzeCompare MeansOne Way ANOVA.Put laboratory into box Factor and t measurement into box Dependent. Under options, select Means Pot Inspect the ANOVA. 1. What is the F test statistic? 2. The F quantile is: F3 os = 2.891564 3. Test the hypothesis. 4. Shown in the SPSS ANOVA is Sig. .It is the p-value. What is it? 5. Use this SPSS p-value to test the hypothesis. 6. Is the quality the same for these four laboratories? 7. Include a paper copy of your SPSS output.

Answer 1

One-way ANOVA: 1, 2, 3, 4 Source DB Factor 3 0.6426 0.2142 7.93 0.000 Error 33 Total 36 1.5342 MS F 0.8916 0.0270 S-0.1644 R-Sq 41.88% R-Sq (adj ) -36.60% - Individual 95% CIs For Mean Based on Pooled StDev Level N Mea StDev 12 0.3725 0.1565 8 0.0490 0.1230 70.0795 0.1570 10 0.1866 0.2024 *-) 0.45 0.00 0.15 0.30 Pooled StDev-0.1644

Group 2 0.00290 0.00739 0.35320 0.00317 0.00543 0.00065 0.00776 0.01149 Group 3 0.10617 0.42503 0.00713 0.00189 0.00515 0.00486 0.00636 Group 1 0.01434 0.10511 0.27031 0.45811 0.45734 0.45710 0.45698 0.45576 0.45466 0.44371 0.45342 0.44324 Group 4 0.21726 0.36502 0.42656 0.00259 0.00397 0.42211 0.00490 0.41975 0.00190 0.00188 4.47008 0.373 1.9344798732 0.156 0.26934527266 0.39199 0.049 0.1250454561 0.123 0.10583843608 0.55659 0.08 0.1920675705 0.157 0.14781150934 1.86594 0.187 0.7168154796 0.202 0.36864227124 Sum- Average- St. Dev. 667 75 286 12 10 n E

The total sample size is N 37. Therefore, the total degrees of freedom are dftotal 37-1-36 Also, the between-groups degrees of freedom are dfbeturen 4 1 degrees of freedom ar 3, and the within-groups dfwithin- dftotal - dfbetween 36-3 33 First, we need to compute the total sum of values and the grand mean. The following is obtained X 4.47008+0.39199+0.55659 + 1.86594 7.2846 1.0 Also, the sum of squared valu es is X1.9344798732 0.1250454561 0.1920675705+0.7168154796 2.9684083794 1.J Based on the above calculations, the total sum of squares is computed as follows 7.28462 ( x, ) -2.9684083794 yx3 1.534 2.J 1.7 The within sum of squares is computed as shown in the calculation below: SuithinSuihngop0.26934527266667 +0.1058384360875+0.147811509342860.36864227124 The between sum of squares is computed directly as shown in the calculation below: 0.892

= 0.214 between - dfbetueen3 MSwihin Shn0.92 feithin330.027 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows FMSetree 0.214 Swithin 0.027 7.927 The following null and alternative hypotheses need to be tested Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA (2) Rejection Region Based on the information provided, the significance level is α = 0.05, and the degrees of freedom are dfi 3 and df 3, therefore, the rejection region for this F-test is R-IF:F>Fe- 2.892) (3)_Test Statistics FM Steen0.214 Swithin -0,027 = 7.927 Since it is observed that F rejected 7.927>F 2.892, it is then concluded that the null hypothesis is

Using the P-value approach: The p-value is p 0.0004, and since p 0.00040.05, it is concluded that the null hypothesis is rejected. () CAnitsi It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 4 population means are equal, at the α-0.05 significance level.