In: Statistics and Probability
2. A food processing plant typically contain fungus spores. If the ventilation system is not adequate, this can have a serious effect of the health of employees. To determine the amount of spores present, random air samples are pumped to a certain plate, and the number of "colony-forming units (CFUs)" are determined after time allowed for incubation. The data from the room of a plant that slaughters 35,000 turkeys per day, which are obtained during the four seasons of the year, is given below. The units are in CFUs per cubic meter.
Fall Winter Spring Summer
1231 384 2105 3175
1254 104 701 2526
1088 97 842 1090
1124 401 1243 1987
a) Examine the data using exploratory data analysis tools. Create at least one graph comparing means.
b) Perform an -way ANOVA to determine is the effect of the season is statistically significant. Use the four-step method. Be sure to give a practical conclusion. Assume the populations are normally distributed and the variances are roughly equal. Copy and paste the results of the test into your Word document.
Load the data into Excel.
Go to Data>Megastat.
Select the option Analysis of Variance and go to One Factor.
Select the data set for the Input Range.
Click OK.
The output obtained will be as follows:
Mean | n | Std. Dev | |||
1,174.3 | 4 | 80.72 | Fall | ||
246.5 | 4 | 168.75 | Winter | ||
1,222.8 | 4 | 631.39 | Spring | ||
2,194.5 | 4 | 882.09 | Summer | ||
1,209.5 | 16 | 865.30 | Total | ||
ANOVA table | |||||
Source | SS | df | MS | F | p-value |
Treatment | 7,596,048.50 | 3 | 2,532,016.167 | 8.36 | .0029 |
Error | 3,635,195.50 | 12 | 302,932.958 | ||
Total | 11,231,244.00 | 15 | |||
Post hoc analysis | |||||
p-values for pairwise t-tests | |||||
Winter | Fall | Spring | Summer | ||
246.5 | 1,174.3 | 1,222.8 | 2,194.5 | ||
Winter | 246.5 | ||||
Fall | 1,174.3 | .0345 | |||
Spring | 1,222.8 | .0275 | .9029 | ||
Summer | 2,194.5 | .0003 | .0223 | .0281 | |
Tukey simultaneous comparison t-values (d.f. = 12) | |||||
Winter | Fall | Spring | Summer | ||
246.5 | 1,174.3 | 1,222.8 | 2,194.5 | ||
Winter | 246.5 | ||||
Fall | 1,174.3 | 2.38 | |||
Spring | 1,222.8 | 2.51 | 0.12 | ||
Summer | 2,194.5 | 5.01 | 2.62 | 2.50 | |
critical values for experimentwise error rate: | |||||
0.05 | 2.97 | ||||
0.01 | 3.89 |
a) Examine the data using exploratory data analysis tools. Create at least one graph comparing means.
The data from the output is:
Mean | n | Std. Dev | |
1,174.3 | 4 | 80.72 | Fall |
246.5 | 4 | 168.75 | Winter |
1,222.8 | 4 | 631.39 | Spring |
2,194.5 | 4 | 882.09 | Summer |
1,209.5 | 16 | 865.30 | Total |
The graph is as follows for the above data comparing means:
b) Perform an -way ANOVA to determine is the effect of the season is statistically significant. Use the four-step method. Be sure to give a practical conclusion. Assume the populations are normally distributed and the variances are roughly equal. Copy and paste the results of the test into your Word document.
The hypothesis of testing is:
Null hypothesis: The effect of the season is not statistically significant.
Alternative hypothesis: The effect of the season is statistically significant.
The ANOVA table from the output is:
ANOVA table | |||||
Source | SS | df | MS | F | p-value |
Treatment | 7,596,048.50 | 3 | 2,532,016.167 | 8.36 | .0029 |
Error | 3,635,195.50 | 12 | 302,932.958 | ||
Total | 11,231,244.00 | 15 |
Since the p-value (0.0029) is less than the significance level, we can say that the effect of the season is statistically significant.
Or
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 7596049 | 3 | 2532016 | 8.358338 | 0.002865 | 3.490295 |
Within Groups | 3635196 | 12 | 302933 | |||
Total | 11231244 | 15 |
Decision & Conclusion:
Since Fcalculated > Fcritical, we can reject the null hypothesis and can say that the effect of the season is statistically significant.