Question

The lengths of service of all the executives employed by Standard Chemicals are:

Answer 1

Solution :

a). There are total 5 executives. Therefore the possible sample size of 2 is

$^nC_r=\frac{n!}{r!(n-r)!}$

$^5C_2=\frac{5!}{2!(5-2)!}$

      $=\frac{5!}{2! \ 3!}$

     = 10

So, there are 10 possible ways for selection of sample size of 2.

b).

     Sample          Samples of service length            Sample mean

Snow, Tolson               20, 22                                    (20+22)/2 = 21  

Snow, Kraft                   20, 26                                         23

Snow, Irwin                   20, 24                                         22

Snow, Jones                 20, 28                                         24

Tolson, Kraft                 22, 26                                         24

Tolson, Irwin                 22, 24                                         23

Tolson, Jones               22, 28                                         25

Kraft, Irwin                     26, 24                                         25

Kraft, Jones                   26, 28                                         27

Irwin,Jones                    24, 28                                         26

c). The mean and the standard deviation of the means of the sampling distribution is given by :

$\bar{X}= \sum_{i-1}^{10}\frac{\bar{x}_i}{n}$

   $=\frac{21+23+22+24+24+23+25+25+27+26}{10}$

   $=\frac{240}{10}$  

  = 24

The variance of the sample means :

$S^2=\frac{1}{n}\sum_{i-1}^{10}\left(\bar x_i - \bar X \right)^2$

   $=\frac{1}{10}\sum_{i-1}^{10}\left(\bar x_i - 24 \right)^2$

  $=\frac{1}{10}\times(30)$

  = 3

Therefore the standard  deviation of the sample means is

$S=\sqrt{variance}$

  $=\sqrt3$

  = 1.732

d). The population means is given by:

$\mu =\frac{20+22+26+24+28}{5}$

  $=\frac{120}{5}$

  = 24

Therefore, we can say that the mean of the sample means is a point estimate of the population mean.