In: Statistics and Probability
Stats I, Item # Q-08
Saleemah, the assistant superintendent in charge of reviewing staffing patterns, is investigating the number of substitute teachers assigned daily to the elementary schools in the district. She ponders whether the numbers vary by school and/or by day of the week. She tests her claims at the 10% significance level, and presumes that the underlying data collection is normally distributed. She gathers independent, simple random samples. The following data represent the numbers of substitute teachers assigned daily to each school during a randomly selected week, cross referenced by school and day of the week:
What are the claims is she testing? What conclusions should she draw? Explain in detail for both factors, both technically and contextually. In particular, if the findings for either one of the two or both factors are especially marginal or strong, please note those results. Cite relevant critical values and p-values that support the findings.
PS#1 |
PS#2 |
PS#3 |
PS#4 |
PS#5 |
PS#6 |
PS#7 |
PS#8 |
|
Mondays |
4 |
8 |
7 |
6 |
2 |
9 |
12 |
8 |
Tuesdays |
5 |
6 |
7 |
4 |
5 |
7 |
10 |
8 |
Wednesdays |
6 |
4 |
7 |
4 |
4 |
5 |
8 |
8 |
Thursdays |
7 |
6 |
7 |
4 |
3 |
3 |
6 |
8 |
Fridays |
3 |
8 |
7 |
6 |
6 |
1 |
4 |
8 |
Anova: Two-Factor Without Replication | ||||||
SUMMARY | Count | Sum | Average | Variance | ||
Mondays | 8 | 56 | 7 | 9.428571429 | ||
Tuesdays | 8 | 52 | 6.5 | 3.714285714 | ||
Wednesdays | 8 | 46 | 5.75 | 3.071428571 | ||
Thursdays | 8 | 44 | 5.5 | 3.714285714 | ||
Fridays | 8 | 43 | 5.375 | 6.267857143 | ||
PS#1 | 5 | 25 | 5 | 2.5 | ||
PS#2 | 5 | 32 | 6.4 | 2.8 | ||
PS#3 | 5 | 35 | 7 | 0 | ||
PS#4 | 5 | 24 | 4.8 | 1.2 | ||
PS#5 | 5 | 20 | 4 | 2.5 | ||
PS#6 | 5 | 25 | 5 | 10 | ||
PS#7 | 5 | 40 | 8 | 10 | ||
PS#8 | 5 | 40 | 8 | 0 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
days | 15.6 | 4 | 3.9 | 1.087649402 | 0.381652762 | 2.7140758 |
PS | 82.975 | 7 | 11.85357143 | 3.305776892 | 0.010850475 | 2.35925985 |
Error | 100.4 | 28 | 3.585714286 | |||
Total | 198.975 | 39 |
claims
not all means at each day are same
not all mean at each PS are same
p-value of days = 0.3817 > 0.10 (alpha)
hence we fail to reject the null hypothesis
p-value of PS= 0.010850475 < 0.10
hence we reject the null hypothesis
we conclude that not all mean at each PS are same