Treść zadania
Autor: venomousdaisy Dodano: 4.2.2014 (17:39)
Statistics Assignment Task 1
A small workshop manufactures steel shafts 25 mm in diameter and 50 mm in length. The critical measurement is the length. A batch of 200 shafts gave the results shown .
The customer will accept the batch if :
the mean value of the batch is within ± 0.03 mm of the nominal value
the spread of values gives a ' Standard Deviation' of less than 0.02 mm
Determine whether this batch will be acceptable.
50.07 50.08 50.08 50.01
50.05 50.04 50.05 50.04
50.10 50.06 50.01 50.04
50.04 50.05 50.06 50.03
50.08 50.04 50.04 50.05
50.01 50.07 50.06 50.03
50.06 50.09 50.06 50.09
50.02 50.07 50.03 50.08
50.04 50.07 50.05 50.02
50.03 50.03 50.04 50.04
50.02 50.04 50.06 50.03
50.05 50.06 50.05 50.07
50.02 50.02 50.08 50.04
50.09 50.03 50.04 50.05
50.05 50.04 50.03 50.05
50.06 50.03 50.05 50.02
50.10 50.01 50.06 50.06
50.06 50.05 50.06 50.01
50.05 50.05 50.05 50.03
50.04 50.04 50.05 50.06
50.08 50.04 50.06 50.00
50.05 50.00 50.08 50.06
50.06 50.08 50.07 50.07
50.09 50.07 50.10 50.04
50.04 50.06 50.05 50.07
50.02 50.10 50.04 50.04
50.03 50.02 50.07 50.05
50.04 50.04 50.06 50.03
50.04 50.05 50.06 50.04
50.05 50.06 50.08 50.05
50.05 50.06 50.07 50.03
49.99 50.06 50.08 50.07
50.04 50.06 50.05 50.08
50.09 50.05 50.06 50.04
50.06 50.08 50.06 50.01
50.04 50.04 50.09 50.05
50.00 50.00 50.04 50.04
50.02 50.06 50.04 50.04
50.02 50.00 50.01 50.07
50.04 50.04 50.02 50.04
50.03 50.05 50.04 50.08
50.04 50.04 50.04 50.07
50.08 50.05 50.09 50.02
50.05 50.03 50.03 50.08
50.06 50.06 50.05 50.06
50.07 50.05 50.04 50.07
50.05 50.01 50.08 50.04
50.05 50.03 50.06 50.05
50.01 50.06 50.06 50.06
50.03 50.06 50.04 50.04
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1 rozwiązanie | autor: venomousdaisy 4.2.2014 (17:38) |
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antekL1 4.2.2014 (18:14)
Please,
take a look at the (included) file: "test.xls", I calculated both values in Excel:
Mean = 50.05 mm - would be acceptable when:
" the mean value of the batch is within ± 0.03 mm of the nominal value"
- ok, the calculated mean is 50.05
so it is 0.05 greater than 50.00, so this batch is NOT acceptable.
Also:
" the spread of values gives a ' Standard Deviation' of less than 0.02 mm"
and calculated value (0.022) is greater than 0.02 too. NOT acceptable.
Remember: the first reason (mean) is enough to reject this batch,
and "standard deviation" is an error of a single measure,
NOT the "error of the mean".
Antek
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