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# One way ANOVA with tukey test calculator

ANOVA is analysis of variance. There are many types of ANOVA test. This calculator is One way ANOVA calculator. ANOVA Table is provided at the end of this solution. The post hoc test we are using is tukey test. the most used post hoc test is Tukey's HSD. Every step is provided as if it is solved by hand. You can learn how to calculate a one-way ANOVA by submitting any sample values. F statistic and the p-value is calculated and shown in Table

INSTRUCTION: Use ',' or new line to separate between values

## You can see a sample solution below. Enter your data to get the solution for your question

$$\displaylines{---}$$
$$\displaylines{}$$
$$Treatment \; no$$
$$1$$
$$2$$
$$3$$

$$1.0$$
$$1.0$$
$$3.0$$

$$2.0$$
$$3.0$$
$$6.0$$

$$3.0$$
$$4.0$$
$$7.0$$

$$4.0$$
$$-$$
$$-$$

$$5.0$$
$$-$$
$$-$$
$$Total$$
$$15.0$$
$$8.0$$
$$16.0$$
$$\displaylines{}$$
$$\displaylines{H_{0}:\;there\;is\;no\;difference\;in\;means \\ \\ H_{a}:\;At\;least\;2\;means\;differ \\ \\ \mathbf{\color{Green}{First\;we\;have\;to\;find\;Total\;mean}} \\ \\ Total\;Mean\;= \frac{Total}{n} \\ \\ \Rightarrow \frac{15.0+8.0+16.0}{11} \\ \\ \Rightarrow \frac{39.0}{11} \\ \\ \Rightarrow 3.545455 \\ \\ \\ \\ \mathbf{\color{Green}{Now\;create\;below\;table\;}} \\ \\ }$$
$$x$$
$$\bar{x}$$
$$x-\bar{x}$$
$$[x-\bar{x}]^{2}$$
$$1.0$$
$$3.545455$$
$$2.545455$$
$$6.479341$$
$$2.0$$
$$3.545455$$
$$1.545455$$
$$2.388431$$
$$3.0$$
$$3.545455$$
$$0.545455$$
$$0.297521$$
$$4.0$$
$$3.545455$$
$$-0.454545$$
$$0.206611$$
$$5.0$$
$$3.545455$$
$$-1.454545$$
$$2.115701$$
$$1.0$$
$$3.545455$$
$$2.545455$$
$$6.479341$$
$$3.0$$
$$3.545455$$
$$0.545455$$
$$0.297521$$
$$4.0$$
$$3.545455$$
$$-0.454545$$
$$0.206611$$
$$3.0$$
$$3.545455$$
$$0.545455$$
$$0.297521$$
$$6.0$$
$$3.545455$$
$$-2.454545$$
$$6.024791$$
$$7.0$$
$$3.545455$$
$$-3.454545$$
$$11.933881$$
$$Total$$


$$36.727271$$
$$\displaylines{}$$
$$\displaylines{\\ \\ From\;the\;table\;we\;can\;get\;SS_{Total} \\ \\ Sum\;of\;square_{Total}= \mathbf{\color{Red}{36.727271}} \\ \\ \mathbf{\color{Green}{Now\;do\;this\;separately\;for\;each\;Treatment}} \\ \\ }$$
$$\displaylines{ \mathbf{\color{Green}{Calculating\;for\;Treatment\;1}} \\ \\ Mean\;of1\;th\;Treatment\;= \frac{Total\;of1\;th\;Treatment\;}{n} \\ \\ \Rightarrow \frac{15.0}{5} \\ \\ \Rightarrow 3.0 \\ \\ }$$
$$x$$
$$\bar{x}$$
$$x-\bar{x}$$
$$[x-\bar{x}]^{2}$$
$$1.0$$
$$3.0$$
$$2.0$$
$$4.0$$
$$2.0$$
$$3.0$$
$$1.0$$
$$1.0$$
$$3.0$$
$$3.0$$
$$0.0$$
$$0.0$$
$$4.0$$
$$3.0$$
$$-1.0$$
$$1.0$$
$$5.0$$
$$3.0$$
$$-2.0$$
$$4.0$$
$$Total$$


$$10.0$$
$$\displaylines{}$$
$$\displaylines{\\ \\ Sum\;of\;square \;of\;Treatment\;No\;1\;= 10.0 \\ \\ \mathbf{\color{Green}{Calculating\;for\;Treatment\;2}} \\ \\ Mean\;of2\;th\;Treatment\;= \frac{Total\;of2\;th\;Treatment\;}{n} \\ \\ \Rightarrow \frac{8.0}{3} \\ \\ \Rightarrow 2.666667 \\ \\ }$$
$$x$$
$$\bar{x}$$
$$x-\bar{x}$$
$$[x-\bar{x}]^{2}$$
$$1.0$$
$$2.666667$$
$$1.666667$$
$$2.777779$$
$$3.0$$
$$2.666667$$
$$-0.333333$$
$$0.111111$$
$$4.0$$
$$2.666667$$
$$-1.333333$$
$$1.777777$$
$$Total$$


$$4.666667$$
$$\displaylines{}$$
$$\displaylines{\\ \\ Sum\;of\;square \;of\;Treatment\;No\;2\;= 4.666667 \\ \\ \mathbf{\color{Green}{Calculating\;for\;Treatment\;3}} \\ \\ Mean\;of3\;th\;Treatment\;= \frac{Total\;of3\;th\;Treatment\;}{n} \\ \\ \Rightarrow \frac{16.0}{3} \\ \\ \Rightarrow 5.333333 \\ \\ }$$
$$x$$
$$\bar{x}$$
$$x-\bar{x}$$
$$[x-\bar{x}]^{2}$$
$$3.0$$
$$5.333333$$
$$2.333333$$
$$5.444443$$
$$6.0$$
$$5.333333$$
$$-0.666667$$
$$0.444445$$
$$7.0$$
$$5.333333$$
$$-1.666667$$
$$2.777779$$
$$Total$$


$$8.666667$$
$$\displaylines{}$$
$$\displaylines{\\ \\ Sum\;of\;square \;of\;Treatment\;No\;3\;= 8.666667 \\ \\ Total\;SS_{Within-treatments}\;=\; 10.0+4.666667+8.666667 \\ \\ \Rightarrow \mathbf{\color{Red}{23.333334}} \\ \\ df_{Between-treatments}=k-1= 3 -1 \\ \\ \Rightarrow 2 \\ \\ df_{Within-treatments}=N-k= 11 - 3 \\ \\ \Rightarrow 8 \\ \\ df_{Total}=N-1= 11 -1 \\ \\ \Rightarrow 10 \\ \\ SS_{Between-treatments} = 36.727271-23.333334 \\ \\ \Rightarrow 13.393937 \\ \\ MS_{Between-treatments}= \frac{SS_{Between-treatments}}{df_{Between-treatments}} \\ \\ \Rightarrow \frac{13.393937}{2} \\ \\ \Rightarrow 6.696968 \\ \\ MS_{Within-treatments}= \frac{SS_{Within-treatments}}{df_{Within-treatments}} \\ \\ \Rightarrow \frac{23.333334}{8} \\ \\ \Rightarrow 2.916667 \\ \\ F = \frac{MS_{Between-treatments}}{MS_{Within-treatments}} \\ \\ \Rightarrow \frac{6.696968}{2.916667} \\ \\ \Rightarrow \mathbf{\color{Red}{2.296103}} \\ \\ p\;value\;is\;\;\mathbf{\color{Red}{0.162912}} \\ \\ \\ \\ \mathbf{\color{Red}{ANOVA\;table\;given\;below}} }$$
$$Sourse$$
$$Sum\;of\;squares$$
$$df$$
$$MS$$
$$F\;test$$
$$p\;value$$
$$Between\;SS$$
$$13.393937$$
$$2$$
$$6.696968$$
$$2.296103$$
$$0.162912$$
$$Within\;SS$$
$$23.333334$$
$$8$$
$$2.916667$$


$$Total$$
$$36.727271$$
$$10$$



$$\displaylines{}$$
$$\displaylines{\alpha\;=\;0.05 \\ \\ p\;is\;more\;than\;\alpha\;.\; So,\;Failed\;to\;Reject\;H_{0} }$$
$$\displaylines{}$$
$$\displaylines{}$$
$$\displaylines{ \mathbf{\color{Red}{Post\;Hoc\;Tukey\;HSD\;test}} \\ \\ \mathbf{\color{Green}{First\;we\;have\;to\;find\;difference\;of\;means\;of\;every\;pair\;possible}} \\ \\ }$$
$$Pairwise\;Comparisons$$
$$Mean\;of\;pair1\;(x_{i})$$
$$Mean\;of\;pair2\;(x_{j})$$
$$(x_{i}-x_{j})$$
$$1nd\;T\;:\;2nd\;T$$
$$3.0$$
$$2.666667$$
$$-0.333333$$
$$1nd\;T\;:\;3nd\;T$$
$$3.0$$
$$5.333333$$
$$2.333333$$
$$2nd\;T\;:\;3nd\;T$$
$$2.666667$$
$$5.333333$$
$$2.666667$$
$$\displaylines{}$$
$$\displaylines{ \mathbf{\color{Green}{Now\;we\;have\;to\;calculate\;confidence\;interval}} \\ \\ \alpha = 0.05 \\ \\ Formula\;for\;confidence\;interval\;=\; x_{i} - x_{j} \pm q*\sqrt{\frac{MS_{within}}{2}(\frac{1}{n_{i}}+\frac{1}{n_{j}}) } \\ \\ \\ \\ Where\;q\;=\; q_{\alpha,k,N-k}\;=\; q_{0.05,3,8}\;=\; 4.037484 \\ \\ MS_{within}\;=\; 2.916667 \\ \\ \Rightarrow x_{i} - x_{j} \pm4.037484\sqrt{\frac{2.916667}{2}(\frac{1}{n_{i}}+\frac{1}{n_{j}}) } \\ \\ \Rightarrow confidence\;interval\;=\; x_{i} - x_{j} \pm4.875725\sqrt{\frac{1}{n_{i}}+\frac{1}{n_{j}} } \\ \\ \mathbf{\color{Green}{Now\;create\;table\;for\;calculating\;confidence\;interval\;using\;above\;formula}} \\ \\ }$$
$$Pairwise\;Comparisons$$
$$(x_{i}-x_{j})$$
$$n_{i}$$
$$n_{j}$$
$$confidence\;interval$$
$$1nd\;T\;:\;2nd\;T$$
$$-0.333333$$
$$5$$
$$3$$
$$-3.89406\;,\;3.227393$$
$$1nd\;T\;:\;3nd\;T$$
$$2.333333$$
$$5$$
$$3$$
$$-1.227393\;,\;5.89406$$
$$2nd\;T\;:\;3nd\;T$$
$$2.666667$$
$$3$$
$$3$$
$$-1.314346\;,\;6.64768$$
$$\displaylines{}$$
$$\displaylines{ \mathbf{\color{Red}{Post\;Hoc\;Tukey\;HSD\;Table}} \\ \\ }$$
$$Pairwise$$
$$x_{i}\;-\;x_{j}$$
$$Confidence\;interval$$
$$p\;value$$
$$Result$$
$$1nd\;T\;:\;2nd\;T$$
$$-0.333333$$
$$-3.89406\;,\;3.227393$$
$$0.9$$
$$Falied\;to\;Reject$$
$$1nd\;T\;:\;3nd\;T$$
$$2.333333$$
$$-1.227393\;,\;5.89406$$
$$0.208377$$
$$Falied\;to\;Reject$$
$$2nd\;T\;:\;3nd\;T$$
$$2.666667$$
$$-1.314346\;,\;6.64768$$
$$0.19668$$
$$Falied\;to\;Reject$$
$$\displaylines{}$$
$$\displaylines{}$$