Question

The EPA questions the report it received from the Zippy Motor Company. The Zippy Company claimed that the motors it has produced are in compliance with the EPA maximum noise emission of 15dB.

Zippy has produced 867 motors and cannot ship them until they are approved by the EPA for noise level compliance. Rather than test each one, they have tested two random samples with results as follows:

Sample A
Size 32

Mean 14.9 dB

Sample B
Size 34

Mean 14.7 dB

The EPA examined several motors and doubted the validity of Zippy’s tests, so it tested the entire lot with the following results:

Entire Lot
Size 867
Mean 15.6 dB Std Dev 0.6 dB

What is the likelihood of the results of the two random samples reported by Zippy explain all steps ? What do you suspect Zippy did to obtain those results?

Answer 1

ANSWER

Given daire Sample B: Sample A nA = 32 Mo - 14.9 dB no = 34 lib147d8 Entire lot Size =867, Mean = 15.6 dB and 5d 0.6dB The probability that the meas EPA of Sample A is greater than or equal to 16.6 dB is calulared as follows: P(XA 3 15.6) = 1-P(TA 2 15.6) 1-0 (1516-MA :) 11-06 15.6-1419 0.61832 21-$ (6.60) - 10.9999 1 ming normidest 0.000) as follows 0 The probability that the Meas EPA of Sample B is greater than con Equal to 15.6 dB is calculared p C#8 >15.6) = l-p KBS 15.6') 15.6 MB Tline 1516-147 1-$(8.75 0161124 .

1 (2 uning nomidist loleulapor) The probabiūty that the mean mean EPA of both the sample exceed the mean EPA of the entire lot is very small. So, Zippy Company will likely to report the resurs of both the samples as the chance of EPÁ more than 3 of getting mean 151.6 is O.