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Cochran q test calculator

Cochran's Q test is a non-parametric statistical test. It is used to test 2 or more treatment have equal effectiveness or not. In Cochran's Q test every input value is either 0 or 1.Every step is provided as if it is solved by hand. You can learn how to calculate a Cochran's Q test by submitting any sample values.


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

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$$ \displaylines{---} $$
$$ \displaylines{ \mathbf{\color{Green}{The\;null\;and\;alternative\;hypothesis\;are:}} \\ \\ \mathbf{\color{Green}{H_{0}:\;the\;treatments\;are\;equally\;effective.}} \\ \\ \mathbf{\color{Green}{H_{a}:\;there\;is\;a\;difference\;in\;effectiveness\;between\;treatments.}} \\ \\ \\ \\ } $$
$$ $$
$$ Treatment\;1 $$
$$ Treatment\;2 $$
$$ Treatment\;3 $$
$$ Block\;total $$
$$ Block\;1 $$
$$ 1 $$
$$ 0 $$
$$ 1 $$
$$ 2 $$
$$ Block\;2 $$
$$ 0 $$
$$ 0 $$
$$ 0 $$
$$ 0 $$
$$ Block\;3 $$
$$ 1 $$
$$ 1 $$
$$ 0 $$
$$ 2 $$
$$ Block\;4 $$
$$ 0 $$
$$ 1 $$
$$ 1 $$
$$ 2 $$
$$ Block\;5 $$
$$ 1 $$
$$ 1 $$
$$ 0 $$
$$ 2 $$
$$ Treatment\;total $$
$$ 3 $$
$$ 3 $$
$$ 2 $$
$$ 8 $$
$$ \displaylines{} $$
$$ \displaylines{K\;=\;Total\;no\;of\;treatment\;=\; 3 \\ \\ N\;=\;Total\;=\; 8 \;(From\;table) \\ \\ \mathbf{\color{Green}{First\;we\;have\;to\;find\;}} \sum_{j=1}^{k}(X_{j}-\frac{N}{k})^{2} \\ \\ \mathbf{\color{Green}{Where\;X_{j}\;is\;jth\;Treatment\;total}} \\ \\ } $$
$$ X_{j} $$
$$ \frac{N}{k} $$
$$ X_{j}-\frac{N}{k} $$
$$ (X_{j}-\frac{N}{k})^{2} $$
$$ 3 $$
$$ 2.666667 $$
$$ 0.333333 $$
$$ 0.111111 $$
$$ 3 $$
$$ 2.666667 $$
$$ 0.333333 $$
$$ 0.111111 $$
$$ 2 $$
$$ 2.666667 $$
$$ -0.666667 $$
$$ 0.444445 $$
$$ Total $$
$$ $$
$$ $$
$$ 0.666667 $$
$$ \displaylines{} $$
$$ \displaylines{ \mathbf{\color{Green}{From\;table\;we\;get\;value\;of\;}} \sum_{j=1}^{k}(X_{j}-\frac{N}{k})^{2} \mathbf{\color{Green}{\;as\;}} \mathbf{\color{Red}{0.666667}} \\ \\ \mathbf{\color{Green}{Now,\;we\;have\;to\;find\;}} \sum_{i=1}^{b}X_{i}*(k-X_{i}) \\ \\ \mathbf{\color{Green}{Where\;X_{i}\;is\;ith\;Block\;total}} \\ \\ } $$
$$ X_{i} $$
$$ k $$
$$ k-X_{i} $$
$$ X_{i}*(k-X_{i}) $$
$$ 2 $$
$$ 3 $$
$$ 1 $$
$$ 2 $$
$$ 0 $$
$$ 3 $$
$$ 3 $$
$$ 0 $$
$$ 2 $$
$$ 3 $$
$$ 1 $$
$$ 2 $$
$$ 2 $$
$$ 3 $$
$$ 1 $$
$$ 2 $$
$$ 2 $$
$$ 3 $$
$$ 1 $$
$$ 2 $$
$$ $$
$$ $$
$$ $$
$$ 8 $$
$$ \displaylines{} $$
$$ \displaylines{ \mathbf{\color{Green}{From\;table\;we\;get\;value\;of\;}} \sum_{i=1}^{b}X_{i}*(k-X_{i}) \mathbf{\color{Green}{\;as\;}} \mathbf{\color{Red}{8}} \\ \\ \mathbf{\color{Green}{The\;Cochran's\;Q\;test\;statistic\;=\;T\;=\;}} k*(k-1) \frac{\sum_{j=1}^{k}(X_{j}-\frac{N}{k})^{2}}{\sum_{i=1}^{b}X_{i}*(k-X_{i})} \\ \\ \Rightarrow 3* 2 * \frac{0.666667}{8} \\ \\ \Rightarrow 0.5 \\ \\ df\;=\; k-1= 2 \\ \\ p\;value\;for\;T\;=\;0.5\;and\;df\;=\;2\;is\; p= 0.778801 \\ \\ p-value\;is\;more\;than\;\alpha \\ \\ So\;There\;is\;not\;enough\;evidence\;to\;reject\;H_{0}\; } $$