In: Statistics and Probability
a.) Write down all feasible samples of size 2 of the values in
the following table and calculate the average number of cases in
each. (Hint: There are 45 possible combinations of size 2.)
b.) Compare the average value of the distribution of sample means
with that of the population. Write what you observe.
c.) In a graph compare the dispersion of the population with that
of the sample means. Write what you observe.
|
Validity |
Zone A |
|
01/01/2019 |
$102.68 |
|
01/01/2018 |
$88.36 |
|
01/01/2017 |
$80.04 |
|
01/01/2016 |
$73.04 |
|
01/01/2015 |
$70.10 |
|
01/01/2014 |
$67.29 |
|
01/01/2013 |
$64.76 |
|
01/01/2012 |
$62.33 |
|
01/01/2011 |
$59.82 |
|
01/01/2010 |
$57.46 |
a)
All the 45 possible combinations of size 2 are shown below . The average is obtained in excel using the function =AVERAGE() and the standard deviation is obtained using the function =STDEV().
| Sample | 1 | 2 | Average |
| 1 | 102.68 | 88.36 | 95.52 |
| 2 | 102.68 | 80.04 | 91.36 |
| 3 | 102.68 | 73.04 | 87.86 |
| 4 | 102.68 | 70.10 | 86.39 |
| 5 | 102.68 | 67.29 | 84.99 |
| 6 | 102.68 | 64.76 | 83.72 |
| 7 | 102.68 | 62.33 | 82.51 |
| 8 | 102.68 | 59.82 | 81.25 |
| 9 | 102.68 | 57.46 | 80.07 |
| 10 | 88.36 | 80.04 | 84.20 |
| 11 | 88.36 | 73.04 | 80.70 |
| 12 | 88.36 | 70.10 | 79.23 |
| 13 | 88.36 | 67.29 | 77.83 |
| 14 | 88.36 | 64.76 | 76.56 |
| 15 | 88.36 | 62.33 | 75.35 |
| 16 | 88.36 | 59.82 | 74.09 |
| 17 | 88.36 | 57.46 | 72.91 |
| 18 | 80.04 | 73.04 | 76.54 |
| 19 | 80.04 | 70.10 | 75.07 |
| 20 | 80.04 | 67.29 | 73.67 |
| 21 | 80.04 | 64.76 | 72.40 |
| 22 | 80.04 | 62.33 | 71.19 |
| 23 | 80.04 | 59.82 | 69.93 |
| 24 | 80.04 | 57.46 | 68.75 |
| 25 | 73.04 | 70.10 | 71.57 |
| 26 | 73.04 | 67.29 | 70.17 |
| 27 | 73.04 | 64.76 | 68.90 |
| 28 | 73.04 | 62.33 | 67.69 |
| 29 | 73.04 | 59.82 | 66.43 |
| 30 | 73.04 | 57.46 | 65.25 |
| 31 | 70.10 | 67.29 | 68.70 |
| 32 | 70.10 | 64.76 | 67.43 |
| 33 | 70.10 | 62.33 | 66.22 |
| 34 | 70.10 | 59.82 | 64.96 |
| 35 | 70.10 | 57.46 | 63.78 |
| 36 | 67.29 | 64.76 | 66.03 |
| 37 | 67.29 | 62.33 | 64.81 |
| 38 | 67.29 | 59.82 | 63.56 |
| 39 | 67.29 | 57.46 | 62.38 |
| 40 | 64.76 | 62.33 | 63.55 |
| 41 | 64.76 | 59.82 | 62.29 |
| 42 | 64.76 | 57.46 | 61.11 |
| 43 | 62.33 | 59.82 | 61.08 |
| 44 | 62.33 | 57.46 | 59.90 |
| 45 | 59.82 | 57.46 | 58.64 |
| Average | 72.59 | ||
| Std.Dev | 9.05 | ||
b.)
similarly for the population, the average is obtained in excel using the function =AVERAGE() and the standard deviation is obtained using the function =STDEV().
| Validity | Zone A |
| 01-01-2019 | 102.68 |
| 01-01-2018 | 88.36 |
| 01-01-2017 | 80.04 |
| 01-01-2016 | 73.04 |
| 01-01-2015 | 70.10 |
| 01-01-2014 | 67.29 |
| 01-01-2013 | 64.76 |
| 01-01-2012 | 62.33 |
| 01-01-2011 | 59.82 |
| 01-01-2010 | 57.46 |
| Average | 72.59 |
| SD | 14.15259 |
We can observe that the mean for the population and all the possible sample is same.
| mean | |
| Population | 72.59 |
| Sample | 72.59 |
c.)
The normal probability plot can be obtained in any online applet. Here i make the plot in Statkey by following these steps,
For population
Step 1: Open StatKey in browser > Theoretical Distribution > Normal. The screenshot is shown below,

Step 2: Click Edit Parameter > enter Mean = 72.59 and Standard Deviation = 14.15259 > OK. The screenshot is shown below,

The normal probability plot is obtained. The screenshot is shown below,

For sample
Step 1: Same
Step 2: Click Edit Parameter > enter Mean = 72.59 and Standard Deviation = 9.05 > OK. The screenshot for the plot is shown below,

Conclusion: From both the plot we can observe that the sample has smaller variance compare to population.