A coladispensing machine is set to dispense 10 ounces of col

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.)

probability=

If the population mean shifts to 9.7, what is the probability that the change will 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 be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

probability=

Solution

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]