IT 223 Assignment #3 Answers
1. Report the average (mean) and standard deviation (sample version) of both variables.
|
|
Position |
Time (ms) |
|
Mean |
6.071429 |
2767.714 |
|
Std Dev |
3.540193 |
1218.608 |
2. Report standard scores for both variables, their products and the correlation coefficient.
This table was created with this Excel file.
|
|
|
|
Position |
Time |
Std Score |
|
|
Position |
Time (ms) |
Std Score |
Std Score |
Product |
|
|
10 |
4727 |
1.109705 |
1.607806 |
1.784191 |
|
|
3 |
4146 |
-0.86759 |
1.131033 |
-0.98127 |
|
|
11 |
3526 |
1.392176 |
0.622256 |
0.866289 |
|
|
7 |
2484 |
0.262294 |
-0.23282 |
-0.06107 |
|
|
1 |
2914 |
-1.43253 |
0.120043 |
-0.17197 |
|
|
8 |
1682 |
0.544764 |
-0.89095 |
-0.48536 |
|
|
8 |
2654 |
0.544764 |
-0.09331 |
-0.05083 |
|
|
9 |
2804 |
0.827235 |
0.029776 |
0.024632 |
|
|
8 |
2934 |
0.544764 |
0.136455 |
0.074336 |
|
|
3 |
1723 |
-0.86759 |
-0.8573 |
0.743784 |
|
|
10 |
4777 |
1.109705 |
1.648837 |
1.829723 |
|
|
2 |
1272 |
-1.15006 |
-1.2274 |
1.411577 |
|
|
3 |
731 |
-0.86759 |
-1.67134 |
1.450039 |
|
|
2 |
2374 |
-1.15006 |
-0.32309 |
0.371567 |
|
Mean |
6.071429 |
2767.714 |
|
Correlation |
0.523511 |
|
Std Dev |
3.540193 |
1218.608 |
|
|
|
The correlation is calculated by
summing the product of the standard scores and dividing by n - 1. You can check
this value by using the CORREL function in excel.
3. Here’s a scatter plot of the data created from Excel. It has the equation for the regression line in the upper right-hand corner. R2 is the square of the correlation.
4. The slope of the regression line can be taken from the equation produced by Excel. It is 180.2.
5. Also from the equation: 1673.6
6. The point of means (6.1, 2767.7) is at the intersection of the standard deviation line and the regression line.
7. This can be calculating by dividing the standard deviation of the y-variable by the standard deviation of the x-variable, which is approximately 344.
8. To predict the time for the fourth variable, simply plug in 4 into the value of x in the regression equation: 180.2 * 4 + 1673.6 = 2394.4ms (or approximately 2.4 seconds). You can verify this amount with the graphed regression line.
9. Approximately 180 milliseconds. The slope of the equation provides a good estimate of how much time is spent scanning each link since this is the amount of additional time needed for each position level. For example, selecting a link in the third position takes 180 ms more time than selecting a link in the second position.
10. The y-intercept is a constant amount of time regardless of the link position. It provides a good estimate of the fixed-cost of selecting a link. Its estimate is 1673 milliseconds (approximately 1.7 seconds).
11. The correlation of .52 is too high to be a coincidence. There is definitely a trend where the links closer to the bottom of the list take more time. Moreover, the estimates of 180 milliseconds for scanning each link and 1.7 seconds for selecting a link seem plausible. However, it’s clear that not all test participants consistently scanned the links from top to bottom. Otherwise, the correlation would have been closer to 1.