A manufacturer has developed a new process for making lightb
Solution
A)
Note that              
 Margin of Error E = t(alpha/2) * s / sqrt(n)              
 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 =    738.4          
 t(alpha/2) = critical t for the confidence interval =    2.073873068          
 s = sample standard deviation =    38.2          
 n = sample size =    23          
 df = n - 1 =    22          
 Thus,              
 Margin of Error E =    16.51891873          
 Lower bound =    721.8810813          
 Upper bound =    754.9189187          
               
 Thus, the confidence interval is              
               
 (   721.8810813   ,   754.9189187   ) [ANSWER]
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b)
We are 95% confident that the true mean is between 721.8810813 and 754.9189187. [ANSWER]
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c)
That the disitribution of the sample means is approximately normal.
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d)
Yes, because 750 is within the confidence interval.

