A tool manufacturer tests three types of cutters used in a lathe operations. One

Question

A tool manufacturer tests three types of cutters used in a lathe operations. One type isa laminated steel cutter consisting of a very hard high carbon steel sandwiched between two softer steels. Another is a special high-speed tool steel cutter developed using powder metallurgy. The final one is made from cryogenically treated steel. Several cutters of each type are tested to see how long they will last (in hours) in continuous operation until they need to be sharpened. The times are recorded for each cutter used in the study, and the results are summarized below. n is the number of cutters used of the type indicated. Average time Type of cutter laminated steel high-speed steel cryogenically treated steel nt 29.84 28.58 33.83 7.08 8.07 6.93 32 27 38 Assume the data are three independent SRSs, one from each of the three populations of cutters, and that the distribution of the times until sharpening is needed is normal A partial ANOVA table is given below Source Cutter type Sums of squares Mean square E-ratio 505.26 53.45 1. The degrees of freedom for cutter type (group) are A) 2 ?) ? C) 94 D) 97 2. The degrees of freedom for error are ?) ? C) 94 D) 97

Explanation / Answer

We have three treatments here types of steel cutter.

1) The degrees of freedom for steel cutter is,

= k - 1

= 3 - 1

= 2

2) The degrees of freedom for Error is,

= N - k

= 97 - 3

= 94

3) This is the randomized designed experiment.

Because it contained randomly data of three treatments no block's there

and not different levels.

ANOVA Table:

MSSt = SSt / 2

= 505.26 / 2

= 252.63

SSE = MSE * df

= 53.45 * 94

= 5024.3

F = MSSt / MSSe

= 252.63 / 53.45

= 4.72647

5) None of the above,

because,

H0: The average time required to sharpening all steel cutter is the same.

against,

H1: at least one cutter needed more time to sharpen.

6) From ANOVA,

F = 4.73 (rounded value to next digit)

7) from above ANOVA table,

p-value = 0.011069

from the p-value we accept the null hypothesis at 5% level of significance

because p-value > 0.05 (l.o.s.)

i.e. The average time required to sharpening all steel cutter is the same.

>>>>>>>>>>>>> Best Luck >>>>>>>>>>>>>

Source df sum of squares Mean square F - ratio p-value Cutter type 2 505.26 252.63 4.72647 0.011069 Error 94 5024.3 53.45 Total 96

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