Copper Spindle
Statistics and Probability?

An article in Electronic Components and Technology Conference (2001, Vol. 52, pp. 1167-1171) compared single versus dual spindle saw processes for copper metallized wafers. A total of 15 devices of each type were measured for the width of the backside chipouts, xbar single = 66.179, s single = 7.894, xbar double = 44.282, s double = 8.841

(a) Do the sample data support the claim that both processes have the same chip outputs? Use alpha = 0.05 and assume that both populations are normally distributed and have the same variance. Find the p-value for the test.

(b) Construct a 95% two-sided confidence interval on the mean difference in spindle saw process.

Let U1 = the true or population mean of the chip outputs in. the single spindle saw processes and U2 = the true or population mean of the chip outputs in the dual spindle saw processes.

(a). We wish to test
Ho: U1-U2 = 0
versus
Ha: U1-U2 ≠ 0
based on the analysis of our sample data.

sp^2 = [(n1-1)*s1^2 + (n2-1)*s2^2]/(n1+n2-2)
= {(15-1)*[(7.894)^2 + (44.282)^2]} /(15*2-2)
= 1011.60538

The test static
= t
= [(x1-bar-x2-bar)-D0]/√[sp^2(1/n1+1/n2)]
= [(66.179-44.282)-0]/√[1011.60538*(1/15+1/15)]
= 1.885426839

Degrees of freedom
= n1+n2-2
= 15*2-2
= 28

Using alpha=0.05 level of significance based on 28 degrees of freedom, then the rejection point
= t(0.05/2);28d.f.
= t0.025;28d.f.
= 2.048

The p-value
= twice the area under the t distribution normal curve to the right of the test static |t|=1.88 at 28 degrees of freedom
= 2*0.0375
= 0.075

Rejection criteria: We can only reject Ho in favor of Ha if and only if |t|>t(alpha/2);28d.f. or the p-value is smaller than that of our chosen alpha value in this case.

Because |t| = 1.88 is smaller than instead of greater than that of t0.025;28d.f. = 2.048, then we cannot or fail to reject Ho in favor of Ha at the 0.05 level of significance and conclude that both processes have the same mean of chip outputs, based on data obtained from our two given independent samples from two normally distributed populations having equal variances. In addition, since the two-tailed p-value is greater than instead of smaller than that of our chosen alpha= 0.05 level of significance, so in other words we don't have sufficient evidence to against Ho and in favor of Ha.

(b).
A 95% two-sided confidence interval on the mean difference in spindle saw process
= [(x1-bar-x2-bar) (+-) t(alpha/2);(n1+n2-2)d.f. * √[sp^2(1/n1+1/n2)]
= {(66.179-44.282)(+-) t0.025;28d.f.*√[1011.60538*(1/15+1/15)]}
= [21.897 (+-) 2.048*√(1011.60538*2/15)]
= [-1.889, 45.68]

The above interval we have got from above tells us that we are 95% confident the mean difference in the width of the backside chipouts in between the single and dual spindle saw processes is between -1.889 and 45.68 units.

Hope this helps.



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