Respuesta :
Answer:
(a) 95% confidence interval for the true mean order size = Â [ 11972.22 , 37067.80 ]
Step-by-step explanation:
We are given a random sample of 10 shipments of stick-on labels with following order sizes;
12,000, 18,000, 30,000, 60,000, 14,000, 10,500, 52,000, 14,000, 15,700, 19,000
Firstly, Sample mean, [tex]Xbar[/tex] = [tex]\frac{\sum X}{n}[/tex]
 = [tex]\frac{12,000+ 18,000+ 30,000 +60,000+ 14,000+ 10,500+ 52,000+ 14,000 +15,700+ 19,000}{10}[/tex] = 24520
Sample standard deviation, s = [tex]\sqrt{\frac{\sum (X-Xbar)^{2} }{n-1} }[/tex] = 17541.81
The pivotal quantity for confidence interval is given by;
      P.Q. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
So, the 95% confidence interval for true mean order size is given by;
P(-2.262 < [tex]t_9[/tex] < 2.262) = 0.95
P(-2.262 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.262) = 0.95
P(-2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < 2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(Xbar - 2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Xbar + 2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ Xbar - 2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] , Xbar + 2.262 * [tex]\frac{s}{\sqrt{n} }[/tex] ]
                      = [ 24520 - 2.262*[tex]\frac{17541.81}{\sqrt{10} }[/tex] ,  24520 - 2.262*[tex]\frac{17541.81}{\sqrt{10} }[/tex] ]
                      = [ 11972.22 , 37067.80 ]
