Consider normal population, N (µ, σ²)

• Set up H₀ : σ²
• Set up H_{1 :} σ²<σ₀² or σ²>σ₀², or σ²≠ σ₀²
• Statistics:

X² =

Here,

S² = Sample Variance (Unbiased)

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

• Set up level of significance α, critical point, X²_tab using the chi-square table along with (n – 1) degree of freedom.
• Computing statistics i.e., X²_cal

H1	Reject H0, if
σ2< σ02	X2cal<X2tabi.e., X21 – α, n -1
σ2> σ02	X2cal> X2tab i.e., X2α, n -1
σ2 ≠ σ02	X2cal< X2tab i.e., X21 – α/ 2, n -1 or,X2cal> X2tab i.e., X2α/ 2, n -1

 

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., S² = 0.000324. take the level of significance as 0.05.

Solution:

• H₀ : σ = 0.022
• H_{1 :} σ ≠ 0.022
• Statistics:

X² =

Here,

S² = 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

X²_{0.025, 17}= 30.191

• Computation

X²_cal =

• Decision:

We have seen that

X_cal< 30.191

X²_cal> 7.564

Thus, it will lie in the accepted region

Therefore,

H₀ is accepted and true diameter of wire rope is 0.022 inch.