Random samples of batting average from the leaders in both l

Random samples of batting average from the leaders in both leagues prior to the All-Star break are shown. At = 0.05, can a difference be concluded? Use the ANOVA test. 6. National 360654 652 338 313 309 American 340 332 317 316 314 306 a) Calculate Total SS, SSE and SST. b) Calculate the value of test statistic F = c) Find the p-value. d) Draw the conclusion National 360 654 652338 313 309 MST MSE

Solution

TSS=X²-[(T²)/n]=1.9085-1.726=.1825. SST=[T_i²/n_i]-[(T²)/n]=[1.1493+.6176]-1.726=.0409. SSE=TSS-SST=.1825-.0409=.1416. F=MST/MSE=(SST/1)/(SSE/10)=.0409/.01416=2.8884. And table value of F (1,10) at 0.05% is 4.96, so Fcal<Ftab, we accept null hypothesis. Therefore, there is no difference in means of the leaders of national and american.

  

TSS=X²-[(T²)/n]=1175200-1038563.636=136636.364. SST=[T_i²/n_i]-[(T²)/n]=[1066107.143-1038563.636=27543.5068. SSE=TSS-SST=136636.364-27543.5068=109092.8572. F=MST/MSE=(SST/2)/(SSE/19)=(27543.5068/2)/(109092.8572/19)=13771.7534/5741.7293=2.3985. And table value of F (2,19) at 0.05% is 3.52, so Fcal<Ftab, we accept null hypothesis. Therefore, there is no difference in mean sodium amounts exist among condiments, cereals and desserts.

National	american
.360	.340
.654	.332
.652	.317
.338	.316
.313	.314
.309	.306
2.626	1.925