∑x = 159.9 , ∑x² = 2557.29
n = 10
Variance, s² = (Ʃx² - (Ʃx)²/n)/(n-1) = (2557.29-(159.9)²/10)/(10-1) = 0.0543
Null and alternative hypothesis:
Hₒ : σ² = 0.2
H₁ : σ² ≠ 0.2
Test statistic:
χ² = (n-1)s² / σ² = (10 - 1)0.0543/0.2 = 2.4450
Degree of freedom:
Df = n -1 = 10 - 1 = 9
p-value = 2*(1-CHISQ.DIST.RT(2.445, 9)) = 0.0354
Decision:
p-value < 0.05, Reject the null hypothesis.
There is enough evidence to conclude that the bag filling machine filling the bag is not in the tolerance limit.
The machinery need adjusting.
4. * (5 points). This sample of 10 data points are taken from quality-control work at...