Question

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A company that manufactures pavers used in residential landscaping. The company guarantees that the paver average weight is no more than 10kg. However, recently the company has received many complaints that the pavers are heavier than expected. In an effort to address the customers complaints, the quality control manager selected a random sample of 40 pavers. The average and standard deviation obtained from this sample are 10.3kg and 3.1 kg, respectively. At α = 0.10, can it be concluded the pavers average weight is higher than 10kg? Use the critical value/rejection point method.

Answer 1

Given : - u (mean) = 10.3 kg ,     (standard deviation) = 3.1 kg and n = 40

Let X be the average weight of the paver i.e. X N ( u , 2 ) ( 10.3 , 3.12 )

Hypothesis:

Ho : = 10

H1 : > 10

Level of Significance (l.o.s.) : = 0.1

Test Statistics : Single mean test for normally distributed data.

Decision Criteria : Reject Ho at 10% l.o.s. if mod ( Zcal  )> Ztab , where  Ztab = Z/2  = Z0.05 = 1.6449 (from Z-table) and Zcal = ( u - ) / (/ )

Calculations :-  

Zcal = ( 10.3 - 10 ) / ( 3.1 / ) = 0.61205374

Conclusion : - Since, mod ( Zcal  ) < Ztab  we do not reject Ho at 10% l.o.s. and thus, conlude that the pavers average weight is 10 kg.