A cola-dispensing machine is set to dispense 10 ounces of cola per cup, with a s

Question

A cola-dispensing machine is set to dispense 10 ounces of cola per cup, with a standard deviation of 0.9 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 35, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit. At what value should the control limit be set? (Round z values to two decimal places. Round your answers to 2 decimal places.) Value should be in between 9.75 and 10.25 If the population mean shifts to 9.7, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.) Probability If the population mean shifts to 10.6, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.) Probability

Explanation / Answer

a)

This is then a 90% confidence level.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    10          
z(alpha/2) = critical z for the confidence interval =    1.64          
s = sample standard deviation =    0.9          
n = sample size =    35          
              
Thus,              
Margin of Error E =    0.249489536          
Lower bound =    9.750510464          
Upper bound =    10.24948954          
              
Thus, the confidence interval is              
              
(   9.75   ,   10.25   )

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    9.75      
x2 = upper bound =    10.25      
u = mean =    9.7      
n = sample size =    35      
s = standard deviation =    0.9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0.33      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3.62      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.6293      
P(z < z2) =    0.9999
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.3706 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    9.75      
x2 = upper bound =    10.25      
u = mean =    10.6      
n = sample size =    35      
s = standard deviation =    0.9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -5.59      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    -2.3      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0      
P(z < z2) =    0.01072411      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.0107 [ANSWER]

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