Flower Color | Flat | Curled |

Pink | 2 | 30 |

Red | 9 | 12 |

Examine the vaccine has a effect in controlling the disease

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Login Now**Solution:**

**Problem to test:**

**Null Hypothesis(H _{0}):** There has no effect in controlling the disease

**Alternate Hypothesis(H _{1}):** There has effect in controlling the disease

**Test Statistics:**

Since, this is the case of 2 by 2 table and one shell frequency is less than 5 then we have formula,

\(X_{cal}^2 = \frac{N[(ad – bc) – \frac{N}{2}]^2}{(a+b)(c+d)(a+c)(b+d)}\)

For calculation of X^{2}

Flower Color | Flat | Curled | Total |

Pink | 2_{ (a)} |
30_{ (b)} |
32_{ (a+b)} |

Red | 9_{ (c)} |
12 _{(d)} |
21 _{(c+d)} |

Total | 11_{ (a + c)} |
42_{ (b + d)} |
53 _{(a+b+c+d = N)} |

Now,

\(X_{cal}^2 = \frac{53.[(2 \times 12 – 30 \times 9) – \frac{53}{2}]^2}{32 \times 21 \times 11 \times 42}\)

\(= \frac{3935581.25}{310464}\)

= 12.67

**Critical Value:** The tabulated value of X^{2} at 0.05 level of significance with (2-1)(2-1) = 1 d.f. is \(X_{(0.05, 1)}^2 = 3.841\)

**Decision:** Since \(X_{cal}^2\) = 12.67 > \(X_{tab}^2\) = 3.841. So, H_{1} is accepted and H_{0} is rejected.

**Conclusion:** Hence, There has effect in controlling the disease.

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