In: Statistics and Probability
Apply the techniques of two-way ANOVA to a data sample split into eight treatments with total degrees of freedom, df = 47, testing at the 2.5% level of significance. The Sum of Squares for Treatments is 42; the Mean Square for Blocks is 9, and the Mean Square for Error is 3.
SS | df | MS | F test Stats | |
Treatments | #1 | #5 | #9 | #12 |
Blocks | #2 | #6 | #10 | #13 |
Error | #3 | #7 | #11 | XXXXXXX |
Total | #4 | #8 | XXXXXXX | XXXXXXX |
#14: What is the F critical for treatments, rounding to the nearest ten-thousandths? [COMMENTS & HINTS: It is a number between zero and ten. Enter this answer as #.####]
#15: What is the F critical for blocks, rounding to the nearest ten-thousandths? [COMMENTS & HINTS: It is a number between zero and ten. Enter this answer as #.####]
#16: What is the most appropriate technical conclusion for treatments, based on the available evidence and testing at the indicated level of significance? {Select the best response.}
a) The researcher would highly reject the unidentified null hypothesis. b) The researcher would reject the unidentified null hypothesis. c) The researcher would marginally reject the unidentified null hypothesis. d) The researcher would highly fail to reject the unidentified null hypothesis. e) The researcher would fail to reject the unidentified null hypothesis. f) The researcher would marginally fail to reject the unidentified null hypothesis.
#17: What is the most appropriate technical conclusion for blocks, based on the available evidence and testing at the indicated level of significance? {Select the best response.}
a) The researcher would highly reject the unidentified null hypothesis. b) The researcher would reject the unidentified null hypothesis. c) The researcher would marginally reject the unidentified null hypothesis. d) The researcher would highly fail to reject the unidentified null hypothesis. e) The researcher would fail to reject the unidentified null hypothesis. f) The researcher would marginally fail to reject the unidentified null hypothesis.
#18: What is the total sample size, n?
#19: What is the p-value, reported to four decimal places, for treatments?
#20: What is the p-value, reported to four decimal places, for blocks?
#21: What is the number of blocks?
SOLUTION
The completed table is,
Source | SS | df | MS | F |
Treatments | 42 | 7 | 6 | 2 |
Blocks | 45 | 5 | 9 | 3 |
Error | 105 | 35 | 3 | |
Total | 192 | 47 |
df for Treatments = Number of treatments - 1 = 8 - 1 = 7
Total number of observations = df Total + 1 = 47 + 1 = 48
Number of blocks = Total number of observations / Number of treatments = 48 / 8 = 6
df for Blocks = Number of blocks - 1 = 6 - 1 = 5
df for Error = df Total - (df treatments + df Blocks) = 47 - (7 + 5) = 35
SS = MS * df for Blocks , Error
SS Total = SS Treatments + SS Block + SS error = 42 + 45 + 105 = 192
MS Treatments = SS Treatments / df Treatments = 42 / 7 = 6
F = MS / MS Error for Treatments and Blocks
14.
For Treatments,
Numerator df = df Treatments = 7
Denominator df = df Error = 35
Critical value of F at df = 7, 35 and 2.5% level of significance is 2.6755
15.
For Blocks,
Numerator df = df Blocks = 5
Denominator df = df Error = 35
Critical value of F at df = 5, 35 and 2.5% level of significance is 2.9557
16.
P-value = P(F > 2, df = 7,35) = 0.0831
Since F for Treatments (2) is less than the critical value of 2.6755, and p-value is under 0.1 but greater than 0.025,
e) The researcher would fail to reject the unidentified null hypothesis
17.
P-value = P(F > 3, df = 5,35) = 0.0234
Since F for Blocks (3) is greater than the critical value of 2.9557, and p-value is just less than 0.025,
c) The researcher would marginally reject the unidentified null hypothesis
18.
Total sample size, n = df Total + 1 = 47 + 1 = 48
19.
P-value for treatments,
P-value = P(F > 2, df = 7,35) = 0.0831
20.
P-value for blocks,
P-value = P(F > 3, df = 5,35) = 0.0234
21.
Number of blocks = 6 (Calculated above)