Users | Brands | |||

Alpha | Beta | Delta | Gamma | |

H1 | A | R | A | R |

H2 | R | A | A | R |

H3 | R | A | R | A |

H4 | A | R | R | R |

H5 | A | A | R | A |

Test whether is any significance difference between brands with respect to acceptability, Use (Cochran Q test)

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Login Now**Solution:**

Lipstick Brands | Users | R_{i} |
R_{i}^{2} |
||||

H_{1} |
H_{2} |
H_{3} |
H_{4} |
H_{5} |
|||

Alpha | A | R | R | A | A | 3 | |

Beta | R | A | A | R | A | 3 | |

Gamma | A | A | R | R | R | 2 | |

Delta | R | R | A | R | A | 2 | |

C_{i} |
2 | 2 | 2 | 1 | 3 | ΣR_{i} = ΣC_{j} = 10 |
ΣR_{i}^{2} = 26 |

C_{i}^{2} |
4 | 4 | 4 | 1 | 9 | ΣC_{i}^{2} = 22 |

Here,

Number of brands (k) = 4

Number of housewives (n) = 5

**Problem to test:**

H_{0}: There is no significant difference between brands.

H_{1}: There is at least one significant difference between brands.

**Test Statistic:**

\(Q = \frac{ (k-1) \left [ k \sum_{i=1}^k R_i^2 – (\sum_{i=1}^k R_i)^2 \right ] }{ k \sum_{j=1}^n C_j – \sum_{j=1}^n C_j^2}\)

\(= \frac{(4-1) (4 \times 26 – (10)^2)}{4 \times 10 – 22}\)

\(= \frac{12}{18}\)

= 0.666

**Critical Value:**

At 5% level of significance then critical value is X^{2}_{(0.05, 3)} = 7.81

**Decision:**

Q = 0.666 < X^{2}_{(0.05, 3)} = 7.81, accept H_{0} at 5% level of significance.

**Conclusion:**

There is no significant difference between brands according to acceptability.

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## Discussion