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Testing of Single Variance

Consider normal population, N (µ, σ2)

  • Set up H0 : σ2
  • Set up H1 : σ202 or σ202, 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, X2tab using the chi-square table along with (n – 1) degree of freedom.
  • Computing statistics i.e., X2cal
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
  • H1 : 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

X20.025, 17= 30.191

  • Computation

X2cal =

  • Decision:

We have seen that

Xcal< 30.191

X2cal> 7.564

Thus, it will lie in the accepted region

Therefore,

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

 

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