1. Central Tendency and Dispersion
sum of x = 42
the sum of x-squared = 262
mean=6
median=6
mode=6
range=4
sample standard deviation = 1.29
population variance=1.43
2. z-scores
-percentile rank Score of 47.5. The z here is +1.5, which cuts off
.5+.4332 of the distribution...so 93.32%tile
-percent of the distribution between 37 and 43. These values
correspond to z=-.6 and z=+.6. .2257 is between the mean and each z,
so the answer is .2257+.2257 or 45.14%
-score cuts off the bottom 10%
of the distribution. The z-table says that z=-1.28 corresponds to the
bottom 10%. Using the z-score formula backwards, that gives
-1.28(5)+40, or 33.6
95% CI with N=30: df=29, t=2.045, so 14.25 and 15.75 fall around mu
mu=13, x-bar=19, s=3, N=51, df=50, critical-t = 2.009
mu=12.5, x-bar=11 s=2, N=61, df=60, critical-t = 2.00
Group 1:
Group 2:
For the entire sample of 8:
Finally, the standard deviation of the means (s-xbar) is 2.298
With the heuristic formula, you'd get:
[4(2.2982]/[(1.412+.52)/2]=18.86
With the raw score formula, you'd get:
3. Confidence interval around mu
95% CI with N=121: df=120, t=1.98, so 14.64 and 15.36 fall around mu
99% CI with N=121: df=120, t=2.617, so 14.52 and 15.48 fall around mu
4. 1-sample t-test
(19-13)/3/51.5 = 14.28
Reject the null hypothesis: Her unit's average sick days are different
from that of the entire company. Specifically, they are worse.
(11-12.5)/2/61.5= -5.858
Reject the null hypothesis: Her unit's average sick days are different
from that of the entire company. Specifically, her unit has a better
record with respect to sick days
5. One-way ANOVA
x = 16,
x2 = 70, mean=4,
s=1.41x = 29,
x2 = 211, mean=7.25,
s=.5x = 45
(i.e., 16+29),
x2 = 281 (ie 70+211).
| Source | Sums of Squares | df | Means Squares | F |
|---|---|---|---|---|
| Total | 281 - 452/8 | 8-1 | ---- | |
| Between | (162+292)/4 - 452/8 | 2-1 | 21.225 | 21.225/1.125 |
| Within | 281-(162+292)/4 | 8-2 | 1.125 |
If you had 3 groups of 4 people, the degrees of freedom would be (3-1) and (12-3), or 2,9
6. Orthogonal Comparisons
Note: The df for a 1-way ANOVA on this problem would be (4-1),(16-4). That would make the MSwithin=11.50/12=.9583
D versus the other groups
[-17 + -14 + -33 +
3(30)]2/
[4[(-1)2+ (-1)2+ (-1)2+32]]
The sums of squares would be 14.083, so F=14.083/.9583 on 1,12 df
A versus B
[17-14]2/
[4[(-1)2+12]]
The sum of squares would be 1.125, so F=1.125/.9583 on 1,12 df
The last set of coefficients would be -1 -1 +2 0, comparing C to B&A
Because the sums of squares for the three contrasts must add up to the
sums of squares between, its SS would be 51.04
7. Bonferroni Comparisons
These comparisons are not mutually orthogonal, so the alpha level must be adjusted to .05/5 = .01 All that changes therefore is the critical F, which is now 9.33. The comparison remains significant.
8. Scheffe Comparisions
D versus the other groups
[-17 + -14 + -33 +
3(30)]2/
[4[(-1)2+ (-1)2+ (-1)2+32]]
The sums of squares would be 14.083. This time, however, the SS does not equal the MS. The
MS is 14.083/3, so F=4.69/.9583 on 3,12 df
A versus B
[17-14]2/
[4[(-1)2+12]]
The mean squares would be 1.125/3, so F=.375/.9583 on 3,12 df
9. Range Tests
First, compute the means for these groups. A=17/4 B=14/4 C=33/4 D=30/4
Second, place them in order from lowest to highest B=3.5 A=4.25 D=7.5 C=8.25
For Tukey, the critical q is 4.20 (4 treatment groups and 12 dferror
8.25-3.5/(.9583/4).5=4.75/.4895=9.704 (Reject null)
8.25-4.25/(.9583/4).5=4/.4895=8.17 (Reject null)
8.25-7.5/(.9583/4).5=.75/.4895=1.53 (Fail to reject null)
7.5-3.5/(.9583/4).5=4/.4895=8.17 (Reject null)
7.5-4.25/(.9583/4).5=3.25/.4895=6.63 (Reject null)
4.25-3.5/(.9583/4).5=.75/.4895=1.53 (Fail to reject null)
Values of obtained q are the same; the critical q changes
8.25 vs 3.5 (Reject null; critical q = 4.20)
8.25 vs 4.25 (Reject null; critical q = 3.77)
8.25 vs 7.5 (Fail to reject null; critical q = 3.08)
7.5 vs 3.5 (Reject null; critical q = 3.77)
7.5 vs 4.25 (Reject null; critical q = 3.08)
4.25 vs 3.5 (Fail to reject null; critical q = 3.08)
10. Two way ANOVA
| Source | Sums of Squares | df | Means Squares | F |
|---|---|---|---|---|
| Total | 439 - 672/12 = 64.92 | 12-1 | ---- | |
| Between | (72+142+212+252)/3 - 672/12 = 62.92 | 2(2)-1 | ------ | |
| Mood | (212+462)/6 - 672/12 = 52.083 | 2-1 | 52.083 | 52.083/.25 |
| Alcohol | (282+392)/6 - 672/12 = 10.083 | 2-1 | 10.083 | 10.083/.25 |
| MoodxAlcohol | 62.92-52.083-10.083 = .754 | (2-1)(2-1) | .754 | .754/.25 |
| Within | 64.92-62.92 = 2 | 12-4 | .25 |
48 participants among the 6 cells of a 2x3 design:
Mood 1,42
Alcohol 2,42
MxA 2,42