Answer:
The value is [tex]UCL = 10.8[/tex] Â
Step-by-step explanation:
From the question we are told that
  The sample mean is  [tex]\= x = 9 \ ounce[/tex]
  The sample size is  n =  25
  The standard deviation is  [tex]\sigma = 3 \ ounce[/tex]
Given that the sample size is not large enough i.e  n<  30  we will make use of the student t distribution table Â
From the question we are told the confidence level is  99.7% , hence the level of significance is  Â
   [tex]\alpha = (100 -99.7 ) \%[/tex]
=> Â [tex]\alpha = 0.003[/tex]
Generally the degree of freedom is  [tex]df = n- 1[/tex]
=> Â [tex]df = 25 - 1[/tex]
=> Â [tex]df = 24[/tex] Â
Generally from the student t distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 24[/tex]  is Â
  [tex]t_{\frac{\alpha }{2} , 24 } = 3.0 [/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E = t_{\frac{\alpha }{2} , 24} *  \frac{\sigma }{\sqrt{n} }[/tex]
=> Â Â [tex]E = 3.0 * Â \frac{3 }{\sqrt{25} }[/tex]
=> Â Â [tex]E =1.8 [/tex]
Gnerally the  upper control chart limit  for 99.7% confidence is mathematically represented as
     [tex]UCL = \= x + E[/tex]
=> Â Â Â [tex]UCL = 9 + 1.8[/tex] Â
=> Â Â Â [tex]UCL = 10.8[/tex] Â
  Â
