Consider normal population, N (µ, σ2)
X2 =
Here,
S2 = Sample Variance (Unbiased)
It follows the chi-square distribution with the (n – 1) degree of freedom.
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:
X2 =
Here,
S2 = Sample Variance (Unbiased)
It follows the chi-square distribution with the (n – 1) degree of freedom.
i.e. α/ 2 = 0.025
X20.025, 17= 30.191
X2cal =
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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