Association

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Assignment 3: Association
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Part A
A research question around two factors that may have an association.. There are a number of product lines handled within the packing department in the company. Within the 10 product lines there are three that typically have the same quantity of work orders shipped at the end of each week. Three packing operators are assigned to pack these three product lines. Each product line has a holding location for the products to be shipped each week. Occasionally when the shipping manager walks through the area he notices that the level of the products in the holding areas can be imbalanced at different days of the week which he believes to be unusual. The objective he set himself was to see if the quantity of the individual product line work orders packed each day, and the day of the week are independent variables.

Hypothesis..
HO: Daily quantity of product lines packed is independent of the day in the week H1: Daily quantity of product lines packed is not independent of the day in the week

Significance Level, α = 0.05

Contingency table..
For a typical week, the manager gathered specific information of the number of work orders packed each day for each line. These observations were tabulated and the results of each variable were calculated and added to the table; Product Line| Mon| Tue| Wed| Thu| Fri| Total|

Zener| 43| 37| 49| 57| 61| 247|
Diode| 46| 48| 39| 45| 42| 220|
Rect| 62| 59| 51| 38| 41| 251|
Total| 151| 144| 139| 140| 144| 718|
Table 1; Contingency Table

Expected Frequency..
The expected frequency table was then calculated. This table gives the expected frequency for each “cell” within the table that would occur if the two variables are completely independent of each other. For each cell, the expected frequency is calculated by;

eij=Row i total(Row j total)Sample Size

 | Mon| Tue| Wed| Thu| Fri| Total|
Zener| 51.95| 49.54| 47.82| 48.16| 49.54| 247|
Diode| 46.27| 44.12| 42.59| 42.90| 44.12| 220|
Rect| 52.79| 50.34| 48.59| 48.94| 50.34| 251|
Total| 151| 144| 139| 140| 144| 718|
Table 2; Expected Frequencies

Test Statistic..
Pearson's chi-squared test uses a measure of goodness of fit which is the sum of differences between observed and expected; each squared and divided by the expectation:

X2=ij(fij-eij)2eij

The following table shows the calculation of each cell

 | Mon| Tue| Wed| Thu| Fri|
Zener| 1.54| 3.17| 0.03| 1.62| 2.65|
Diode| 0.00| 0.34| 0.30| 0.10| 0.10|
Rect| 1.61| 1.49| 0.12| 2.45| 1.73|
Table 3; constructs for test statistic

Giving a resultant test statistic, X2, value of 17.26

Reject or accept the hypothesis?

Rejection rule; reject H0 if p-value ≤ α or X2 ≥ X∝2

Degrees of freedom = (3-1)(5-1) = 8

Critical value from Chi-squared table;
With α = 0.05 and D.F. = 8, X0.052 = 15.507

p-value approach;

X2 = 17.26, gives an area in the upper tail between 0.05 & 0.025

The p-value ≤ α. We can reject the null hypothesis

Critical Value approach

X2 = 17.26 ≥ X0.052 = 15.507. We can reject the null hypothesis

Conclusions
At the 0.05 level of significance, we reject the assumption that the that the daily quantity of product lines packed is independent of the day in the week. Subsequent to this finding, the shipping manager interviewed the operators involved in the packing of these three product lines and found that one of the operators had a higher output than the other two, and one operator had a lower than typical output. The two operators, responsible for the Rectifier and Zener lines respectively, had previously arranged to swop product line responsibilities at midday on Wednesdays to balance the packing rate for the week. This resulted in the relatively high volume of product early in the week in the Zener line and yet all requirements...
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