could solve this question first by hand then use minitab program please An artic

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could solve this question first by hand then use minitab program please

An article describes a study comparing single versus dual spindle saw processes for copper metallized wafers. A total of 13 devices of each type were measured for the width of the backside chip outs, x_single = 65.385 x_double = 55.278 s_single = 6.895 s_double = 7.612 assume alpha = 05 a. Do the sample data support the claim that both processes have the same chip outputs? Assume that both applications are normally distributed and have the same variance b. Construct a 95% two sided confidence interval on the mean difference in spindle saw process. Compare this interval to the results in part (a)

Explanation / Answer

(a)

Data:        

n1 = 13       

n2 = 13       

x1-bar = 65.385       

x2-bar = 55.278       

s1 = 6.895       

s2 = 7.612       

Hypotheses:        

Ho: 1 = 2        

Ha: 1 2        

Decision Rule:        

= 0.05       

Degrees of freedom = 13 + 13 - 2 = 24      

Lower Critical t- score = -2.063898547       

Upper Critical t- score = 2.063898547       

Reject Ho if |t| > 2.063898547       

Test Statistic:        

Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   (((13 - 1) * 6.895^2 + (13 - 1) * 7.612^2)/(13 + 13 -2)) =     7.262

SE = s * {(1 /n1) + (1 /n2)} = 7.26235392279941 * ((1/13) + (1/13)) = 2.848529567      

t = (x1-bar -x2-bar)/SE = 3.548146426       

p- value = 0.001635234       

Decision (in terms of the hypotheses):        

Since 3.548146426 > 2.063898547 we reject Ho and accept Ha    

Conclusion (in terms of the problem):     

There is sufficient evidence that the means are different.   

(b)

n1 = 13

n2 = 13

x1-bar = 65.385

x2-bar = 55.278

s1 = 6.895

s2 = 7.612

% = 95

Degrees of freedom = n1 + n2 - 2 = 13 + 13 -2 = 24

Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((13 - 1) * 6.895^2 + ( 13 - 1) * 7.612^2)/(13 + 13 -2)) = 7.262353923

SE = Pooled s * ((1/n1) + (1/n2)) = 7.26235392279941 * ((1/13) + (1/13)) = 2.848529567

t- score = 2.063898547

Width of the confidence interval = t * SE = 2.06389854731807 * 2.8485295666901 = 5.879076035

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = 10.107 - 5.87907603468426 = 4.227923965

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = 10.107 + 5.87907603468426 = 15.98607603

The 95% confidence interval is [4.23, 15.99]

The confidence interval excludes 0. This means there is a significant difference between the means. The result of Part (a) is verified.

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