Lasers are used to inspect solderjoint defects on printed ci

Lasers are used to inspect solder-joint defects on printed circuit boards (PCB). A manufacturer of a laser-based inspected equipment claims that its product can inspect at least 10 solder-joints per second. But a potential buyer does not agree. Here are the number of solder-joints inspected on each run:

10 9 10 7 12 7 10 7 10 8 8 11 8

1) Does the data support the buyer? State all the hypothesis testing steps and use = 0.01.

2) Find the p-value.

3) Form a 95% confidence interval for the population mean value of solder-joints inspected.

Solution

1.

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   10  
Ha:    u   <   10  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    12          
tcrit =    -   2.680997993      
              
Getting the test statistic, as              
              
X = sample mean =    9          
uo = hypothesized mean =    10          
n = sample size =    13          
s = standard deviation =    1.632993162          
              
Thus, t = (X - uo) * sqrt(n) / s =    -2.207940217          

As |t| < 2.680997, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

Thus, there is no sufficient evidnece to support the buyer\'s claim. [CONCLUSION]

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2.  
              
Also, the p value is              
              
p =    0.023727816   [ANSWER]

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3.

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    9          
t(alpha/2) = critical t for the confidence interval =    2.17881283          
s = sample standard deviation =    1.632993162          
n = sample size =    13          
df = n - 1 =    12          
Thus,              
              
Lower bound =    8.013192109          
Upper bound =    9.986807891          
              
Thus, the confidence interval is              
              
(   8.013192109   ,   9.986807891   ) [ANSWER]