PAIRED NET TYPES, WITH NET TYPE A AND B ON EACH BOAT.
a. 95% CI on mean difference between net types A and B. Recall that once
we have d-bar=0.73 and s_d=0.89, we do a one-sample 95% CI around d-bar:
.73 +- 2.447*(.89/sqrt(7)) 2.447=t.05(2) for 6 df.
(-.093, 1.55) is the final 95% CI for mu-d, the difference between the
catches for the two net types (A-B). Since zero is contained within the
95% CI, it is a plausible, or believable, or "acceptable" result. "No
difference between the two net types" is a credible possibility given the
data, so we cannot reject the null hypothesis that the two net ypes
results in the same mean catch.
b. mu-d=1.0=delta; s^2/delta^2 = 0.7921
n=infinity: t-alpha=1.96, t-beta=0.6745, n>=6
n=6; t-alpha=2.571, t-beta=0.727, n>=9
n=9; t-alpha=2.306, t-beta=0.706, n>=8
n=8; t-alpha=2.365, t-beta=0.711, n>=8
n=8 pairs
c.
Ho: group1-group2=0
T-plus=25; T-minus=3;
25+3=28=7*8/2
T-min=3, T(7,.05(2))=2; 3>2, fail to reject Ho.
DOES THE TYPE B BACTERIAL CULTURE PROMOTE HIGHER BACTERIAL GROWTH THAN
TYPE A CULTURE, USING A NONPARAMETRIC TEST? [5% level of significance]
Ho: Distribution of bacterial growth of the two cultures is the same.
Ha: Type B culture promotes higher bacterial growth than the Type A
culture.
Type A ranks: 3, 3, 3, 3, 6.5
Type B ranks: 3, 6.5, 8, 9, 10.
Sum of ranks for Type A = 18.5. Call this rsA.
Sum of ranks for Type B = 36.5. Call this rsB.
As a check, they add up to 55, which equals 10(11)/2. "Algebraic check".
Because we have a one-sided alternative hypothesis, we need only consider
the rank sum that is expected to be smaller under Ha. Under Ha, we expect
rsB to be larger and rsA to be smaller. [see notes p. 10-13, Zar p. 150].
So the U-observed is based upon U1, which uses rsA:
U-obs = nA*nB + nA(nA+1)/2 - rsA = 5*5 = 5(6)/2 - 18.5 = 21.5. Compare
this with the .05(1) value for nA=nB=5 from Table B.11, which is 21. So we
reject the Ho and conclude that Type B culture promotes higher bacterial
growth than Type A culture. Additionally, .025 < P(1) < .05.
TWO-SAMPLE t-TEST FOR UPSTREAM DISTANCES BY WILD AND HATCHERY STEELHEAD
FISH
a. F-test for equality of the two variances. F-observed = 3.53^2/1.96^2 =
3.24; compare to F-tabled value of 4.82 [df=9,7, .05(2)]. Fail to reject
Ho that the two variances are equal.
b. Ho: mean distances are equal.
Ha: wild steelhead achieve a greater mean distance upstream than the
hatchery fish. alpha(1) = .10.
Use a two-sample, pooled variances, t-test. First, find sp^2, the pooled
variance. sp^2 = [9*3.53^2 + 7*1.96^2]/(7+9) = 8.69.
Second, find denominator of t-obs: sqrt(8.69*(1/10 + 1/8)) = 1.40.
Then t-obs. = (1.95-1.33)/1.40 = 0.44. Compare to .10 1-sided critical
value [16 df] of 1.337. Since t-0bs. < 1.337, we fail to reject the null
hypothesis. There is not a significant difference between the distances
traveled by wild and hatchery salmon. Associated P-value > 0.25.
c. 95% CI for mean difference in distances for the two groups is:
(1.95-1.33) +- 2.120*1.40 = [-2.35, 3.59]. The value of zero
[corresponding to "no difference between the two groups"] is inside this
95% CI. Of course, the confidence interval I really should have asked for,
to be consistent with part (b), is an 80% CI. (alpha(1)=.10 corresponds to
.10 probability in one tail, which corresponds to .20 probability in two
tails, which corresponds to .80 probability in the middle.) In that case,
the 80% CI is (1.95-1.33) +- 1.337*1.40 = [-1.25, 2.49]. This is shorter
than the 95% CI, as would be expected, and still contains the value of
zero, indicating "no difference between the two groups" with respect to
mean distance achieved upstream.
d. Given sp=3.0. The t-alpha value is 1.337 and the t-beta
value is .690. Then delta, the minimum detectable distance between the
two population means, is
(1.337 + .690)*3.0*sqrt(1/10 + 1/8) = 2.88 km is the minimum detectable
difference. Note that 2.88 is larger than the observed difference between
the two sample means. So it makes sense that we did not reject the null
hypothesis of equality between the two population means.