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

Answer:

(a) 95% confidence interval for the true mean order size =  [ 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, $Xbar$ = $\frac{\sum X}{n}$

  = $\frac{12,000+ 18,000+ 30,000 +60,000+ 14,000+ 10,500+ 52,000+ 14,000 +15,700+ 19,000}{10}$ = 24520

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

The pivotal quantity for confidence interval is given by;

            P.Q. = $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

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 < $\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P(-2.262 * $\frac{s}{\sqrt{n} }$ < $Xbar - \mu$ < 2.262 * $\frac{s}{\sqrt{n} }$ ) = 0.95

P(Xbar - 2.262 * $\frac{s}{\sqrt{n} }$ < $\mu$ < Xbar + 2.262 * $\frac{s}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [ Xbar - 2.262 * $\frac{s}{\sqrt{n} }$ , Xbar + 2.262 * $\frac{s}{\sqrt{n} }$ ]

                                           = [ 24520 - 2.262*$\frac{17541.81}{\sqrt{10} }$ ,  24520 - 2.262*$\frac{17541.81}{\sqrt{10} }$ ]

                                           = [ 11972.22 , 37067.80 ]