Consider normal population, N (µ, σ^{2})

- Set up H
_{0}: σ^{2} - Set up H
_{1 : }σ^{2}<σ_{0}^{2}or σ^{2}>σ_{0}^{2}, or σ^{2}≠ σ_{0}^{2} - Statistics:

X^{2} =

Here,

S^{2} = Sample Variance (Unbiased)

It follows the chi-square distribution with the (n – 1) degree of freedom.

- Set up level of significance α, critical point, X
^{2}_{tab}using the chi-square table along with (n – 1) degree of freedom. - Computing statistics i.e., X
^{2}_{cal}

H_{1} | Reject H_{0}, if |

σ^{2}< σ_{0}^{2} | X^{2}_{cal}<X^{2}_{tab}i.e., X^{2}_{1 – α, n -1} |

σ^{2}> σ_{0}^{2} | X^{2}_{cal}> X^{2}_{tab }i.e., X^{2}_{α, n -1} |

σ^{2} ≠ σ_{0}^{2} | X^{2}_{cal}< X^{2}_{tab }i.e., X^{2}_{1 – α/ 2, n -1}or,X |

**Example No. 12:**

Suppose null hypothesis σ = 0.022 inch for certain wire rope diameter. Its alternate hypothesis is against σ ≠ 0.022 inch for a random sample of 18 yields i.e., S^{2} = 0.000324. take the level of significance as 0.05.

**Solution:**

- H
_{0}: σ = 0.022 - H
_{1 : }σ ≠ 0.022 - Statistics:

X^{2} =

Here,

S^{2} = Sample Variance (Unbiased)

It follows the chi-square distribution with the (n – 1) degree of freedom.

- Here, α = 0.05 and the alternative hypothesis will be right tailed test

i.e. α/ 2 = 0.025

X^{2}_{0.025, 17}= 30.191

- Computation

X^{2}_{cal} =

- Decision:

We have seen that

X_{cal}< 30.191

X^{2}_{cal}> 7.564

Thus, it will lie in the accepted region

Therefore,

H_{0} is accepted and true diameter of wire rope is 0.022 inch.