Solution:
Solution:
Confidence interval for population standard deviation is given as below:
Sqrt[(n – 1)*S2 / χ2α/2, n– 1 ] < σ < sqrt[(n – 1)*S2 / χ21 -α/2, n– 1 ]
We assume Confidence level = 95%
From given data, we have
Sample size = n = 5
Degrees of freedom = n – 1 = 4
Sample standard deviation = S = 4.8844
χ2α/2, n – 1 = 11.1433
χ21 -α/2, n– 1 = 0.4844
(By using chi square table)
Sqrt[(5 – 1)* 4.8844^2/ 11.1433] < σ < sqrt[(5 – 1)* 4.8844^2/ 0.4844]
Sqrt[8.5639] < σ < sqrt[196.9979]
2.9264 < σ < 14.0356
Lower limit = 2.9264
Upper limit = 14.0356
Confidence interval does not suggest that the equipment requires adjustment, because the above interval contains the value 5.5.
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