Answer:
The value is [tex]UCL = 10.8[/tex]
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
From the question we are told the confidence level is 99.7% , hence the level of significance is
[tex]\alpha = (100 -99.7 ) \%[/tex]
=> [tex]\alpha = 0.003[/tex]
Generally the degree of freedom is [tex]df = n- 1[/tex]
=> [tex]df = 25 - 1[/tex]
=> [tex]df = 24[/tex]
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
[tex]t_{\frac{\alpha }{2} , 24 } = 3.0 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2} , 24} * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 3.0 * \frac{3 }{\sqrt{25} }[/tex]
=> [tex]E =1.8 [/tex]
Gnerally the upper control chart limit for 99.7% confidence is mathematically represented as
[tex]UCL = \= x + E[/tex]
=> [tex]UCL = 9 + 1.8[/tex]
=> [tex]UCL = 10.8[/tex]
