Question

Problem 3: Two types of drill bits are being tested for wear. The two types of drill bits are each tested on 10 different machines, each used for a variety of parts. The data on wear in mm, over a 2-week testing cycle is shown below. Drill Bit Machine 1 2.6 1 2 3 2.9 3.2 3.0 2.4 1.4 3.4 1.8 3.3 2.4 2.0 2.8 2.7 3.5 3.0 2.7 3.1 3.6 10 2.8 (a) Is there evidence to suggest that the two types of drill bits wear differently? Use a hypothesis test with a = 0.05 to make your conclusions. (b) Is it believable that Drill Bit 2 has significantly less wear than Drill Bit 1? Use a 99% CI to make this decision and state your conclusions.

Answer 1

Solution-

The hypothesis test is T test for two samples is used for this problem and performed using statistical calculator.

Summary and test statistics

Treatment 1 (X) 3.2 3.0 2.4 1.4 3.4 2. 0 Diff (X-M) 0.36 0. 16 -0.44 -1.44 | 0.56 -0.84 -0.04 | 0.66 0. 26 | 0.76 Sq. Diff (X- My2 0.13 0.03 0.19 2. 07 0.31 0.71 0.00 0.44 0.07 0.58 3.5 3. 1 3.6 M : 2.84 SS: 4.52 Treatment 2 (X) 2.6 2.9 2.7 1.8 3.3 Diff (X-M) -0.09 | 0.21 | 0.01 -0.89 | 0.61 -0.29 0.01 | 0.31 0.01. 0.11 Sq. Diff (X-M)2 0.01 0.04 0.00 0.79 0.37 0 . 08 | 0.00 | 0 . 10 0.00 0. 01 ទះ៖ M : 2.69 SS: 1.41

​​​​The t-value is 0.58422. The p-value is 0.566324. The result is not significant at p < 0.05.

Difference Scores Calculations Treatment 1 N1:10 df = N - 1 = 10 - 1 = 9 M1: 2.84 SS1: 4.52 521 = SS1/(N - 1) = 4.52/(10-1) = 0.5 Treatment 2 N2: 10 df2 = N - 1 = 10 - 1 = 9 M2: 2.69 SS2: 1.41 522 = SS2/(N - 1) = 1.41/(10-1) = 0.16 T-value Calculation s?p= ((df/(df, + df2) * 521) + ((dfx/(dfx + df)) * $22) = ((9/18) * 0.5) + ((9/18) * 0.16) = 0.33 SPM, = 52 / N1 = 0.33/10 = 0.03 s2 m2 = 52 p/ N2 = 0.33/10 = 0.03 t= (M4 - M2)/115-M, +52 M2) = 0.15/70.07 = 0.58

Hence, There is no evidence to suggest that the two type of drill bits wear differently.

B)

The t-value is 0.5839. The p-value is .2810. The result is not significant at p < .01.

Significance level: 0.01 Step 1 HO: u1 = u2 H1: u1 > u2 Step 2 Formula Tcal for two means Step 3 CR > 2.552 Step 4 Tcal = 0.5839 Step 5 The value of T calculated does not belong critical region, therefore, HO is not rejected at a 0.01 significance level. P-Value: 0.2810

Hence, Drill Bit 2 has not significantly less than Drill Bit 1.

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