Question

A random sample of 10 shipments of stick-on labels showed the following order sizes. 12,000 18,000 30,000 60,000 14,000 10,500 52,000 14,000 15,700 19,000 Click here for the Excel Data File (a) Construct a 95% confidence interval for the true mean order size. (Round your standard deviation answer to 1 decimal place and t-value to 3 decimal places. Round your answers to the nearest whole number.) The 95% confidence interval to

Answer 1

(a) 95% confidence interval for the true mean order size = [ 11972.22 , 37067.80 ]

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,
`Xbar` =
`(\sum X)/(n)`

=
`(12,000+ 18,000+ 30,000 +60,000+ 14,000+ 10,500+ 52,000+ 14,000 +15,700+ 19,000)/(10)` = 24520

Sample standard deviation, s =
`\sqrt{(\sum (X-Xbar)^(2) )/(n-1) }` = 17541.81

The pivotal quantity for confidence interval is given by;

P.Q. =
`(Xbar - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

So, the 95% confidence interval for true mean order size is given by;

P(-2.262 <
`t_9` < 2.262) = 0.95

P(-2.262 <
`(Xbar - \mu)/((s)/(√(n) ) )` < 2.262) = 0.95

P(-2.262 *
`(s)/(√(n) )` <
`Xbar - \mu` < 2.262 *
`(s)/(√(n) )` ) = 0.95

P(Xbar - 2.262 *
`(s)/(√(n) )` <
`\mu` < Xbar + 2.262 *
`(s)/(√(n) )` ) = 0.95

95% confidence interval for
`\mu` = [ Xbar - 2.262 *
`(s)/(√(n) )` , Xbar + 2.262 *
`(s)/(√(n) )` ]

= [ 24520 - 2.262*
`(17541.81)/(√(10) )` , 24520 - 2.262*
`(17541.81)/(√(10) )` ]

= [ 11972.22 , 37067.80 ]