CHAPTER 07

7.11a: 95%C.I.: mu1 – mu2 = (92 – 115) +/- (2.447)(9.192) (using df = 6)

-45.5 < mu1 – mu2 < -0.5 nmol/gm.

7.11b: 90%C.I.: mu1 – mu2 = (92 – 115) +/- (1.943)(9.192) (using df = 6)

-40.9 < mu1 – mu2 < -5.1 nmol/gm.

7.21: Let 1 denote red and let 2 denote green.

SE_{y1-bar – y2-bar} = sqrt[.36^2 + .36^2] = .509

95% C.I.: (8.36 – 8.94) +/- (2.042)(.509) = -0.58 +/- 1.039 (using df = 30)

or (-1.62,0.46) or -1.62 < mu1 – mu2 < 0.46

7.22: No; the sample sizes are moderately large, so the sampling distribution of the difference in means is approximately normal, despite the skewness in the underlying data.

7.27: H0: muHH = muSH

HA: muHH not= mush

SE_{y1bar – y2bar} = sqrt[ 17.8^2/33 + 19.1^2/51] = 4.09

ts = (18.3 – 13.9)/4.09 = 1.07. Welch’s formula (7.1) yields 71.9, so df = 71.

.20 < P-value < .40. Thus P-value > alpha and we do not reject H0.

7.30a:

H0: mean tibia length does not depend on gender (m1 = mu2)

HA: mean tibia length depends on gender (mu1 not= mu2)

Alpha = .05

SE_{y1bar – y2bar} = sqrt[ 2.87^2/60 + 3.52/50] = .62056

ts = (78.42 – 80.44)/.62056 = -3.26. Welch’s formula (7.1) yields 94.3, so df = 94.

Using Table 4 with df = 80, we obtain .001 < P-value < .01 < alpha, so we reject H0.

7.30b: There is strong evidence (.001 < P-value < .01) to conclude that mean tibia length is larger in female cicada than male cicada.

7.30c: Judging from the means and SDs, the two distributions overlap substantially, so tibia length would be a poor predictor of sex. This is so in spite of the significant difference in female and male tibia lengths.

7.30d: H0 and HA are as in part (a).

SE_{y1bar – y2bar} = sqrt[ 2.87^2/6 + 3.52/5] = 1.962

ts = (78.42 – 80.44)/1.962 = -1.03. Welch’s formula (7.1) yields 7.8, so df = 7.

Using Table 4 with df = 7, we obtain alpha < .20 < P-value < .40, so we retain H0.

There is insufficient evidence (0.20 < P-value < 0.40) to conclude that the tibia lengths of male and female cicadas differ.

7.31: (a) Population 1 = thymus weights at 14 days of all similar chick embryos

Population 2 = thymus weights at 15 days of all similar chick embryos

Here, the null hypothesis is

H0: m1 = m2,

and the alternative is

HA: m1 NE m2, where NE = not equal to.

Since the sample sizes are so small (each under 25), we must assume that each of the populations follow a normal distribution.

Here SE1 = 8.73/sqrt(5) = 3.9042 and SE2 = 7.19/sqrt(5) = 3.2155, so

SEy1-bar - y2-bar = sqrt[3.9042^2 +3.2155^2] = 5.0579 and

Welch's Formula (7.1) is 7.717, so take df = 7. The test statistic is ts = [(31.72 - 29.22) - (0)]/5.0579 = 0.49

and so the P-value is greater than 40% (from Minitab, it is 64%).

Since the P-value is greater than a (10% here), we fail to reject the null hypothesis.

Thus, there is insufficient evidence in this study to reject the claim that the thymus weight at day 14 is equal to the thymus weight at day 15.

(b) The difference is not statistical significant. Thus the fact that y1-bar is greater than y2bar could easily be attributed to chance.

Indeed, a student has commented that it would not be surprising if mu1 were actually less than mu2, because in certain cases the thymus gland would be expected to

shrink during embryonic development.

7.79a: The null and alternative hypotheses are

H0: Toluene has not effect on dopamine in rat striatum

HA: Toluene has some effect on dopamine in rat striatum

Let 1 denote toluene and 2 denote control.

For the K1 count, we note that there are four Y2’s less than the first Y1; there are five Y2’s less than the second Y1; there are five Y2’s less than the third Y1; and there are six Y2’s less than the fourth, fifth, and sixth Y1. Thus

K1 = 4 + 5 + 5 + 6 + 6 + 6 = 32.

For the K2 count, we note that there are no Y1’s less than the first, second, third , and fourth Y2; there is one Y1 less than the fifth Y2; and ther are three Y1’s less than the sixth Y2. Thus

K2 = 0 + 0 + 0 + 0 + 1 + 3 = 4.

To check the counts, we verify that K1 + K2 = 32 + 4 = 36 = (6)(6) = (n1)(n2).

The Wilcoxon-Mann-Whitney test statistic is the larger of the two counts K1 and K2; thus Us = 32.

Looking in Table 6 under n = 6 and n’ = 6, we find that for a nondirectional alternative, the .05 entry is 31 and the .02 entry is 33. Thus, the P-value is bracketed as 0.02 < P-value < 0.05.

At significance level alpha = 0.05, we reject H0, since P-value < 0.05. We note that K1 is larger than K2, which indicates a tendency for the Y1’s to be larger than the Y2’s. Thus there is sufficient evidence (0.02 < P-value < 0.05) to conclude that toluene increases dopamine in rat striatum.

7.79b: When conducting a directional test, we must check directionality. In this case, we note that K1 is larger than K2, which indicates a tendency for the Y1’s to be larger than the Y2’s, which is what the directional alternative predicts. We proceed as in part (a), except that we use the “directional” tail probabilities. Thus, 0.01 < P-value < 0.25. We reject H0 and conclude that there is sufficient evidence (0.01 < P-value < 0.025) to conclude that toluene increases dopamine in rat striatum.

7.80: Here, the null hypothesis is that ventilation is not differently affected by the

"to be hypnotized" and the "control" conditions, whereas the alternative is that

ventilation is indeed differently affected by these two conditions.

First, note that the experimental values are often higher than the control group,

so we really only need find K1. The number of values from Control group less

than 532 is 4, less than 560 is 5, less than 574 is 6, less than 606 is 6, and less

than 632, 634, 679 and 718 is each 8. Thus, K1 is 4+5+6+6+8+8+8+8 = 53. Next,

using Table 6 (since this is a two-tailed test), note that the P-value is between

0.02 and 0.05. Since alpha is 10%, we reject the null hypothesis, and conclude

that ventilation rates are indeed different for the two conditions.

Incidentally, MTB provided a p-value of 0.0313 for these data.

Minitab output (obtained by "Stat", "Nonparametrics", "Mann-Whitney")

Mann-Whitney Confidence Interval and Test

Expl N = 8 Median = 6.190

Control N = 8 Median = 5.135

Point estimate for ETA1-ETA2 is 0.895

91.7 Percent CI for ETA1-ETA2 is (0.240,1.530) W = 89.0

Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0313

7.80c: The t test requires normality for the populations, whereas the Wilcoxon does not (both require independent, random samples). The frequency distribution for the females highly skewed, due to the two large observations of 10.7 and 11.7. This casts doubt on the normality condition.

7.84: Let 1 denote joggers and let 2 denote fitness program entrants.

H0: There is no difference in resting blood concentration of HBE between joggers and fitness program entrants

HA: There is a difference in resting blood concentration of HBE between joggers and fitness program entrants.

K1 = 93.5, K2 = 71.5, Us = 93.5. With n = 15 and n’ = 11, 108 is under the .20 heading for a nondirectional (two tailed) alternative and is the smallest entry listed. Thus Us = 93.5 is off the chart, which means that P-value > .20 and H0 is retained. There is insufficient evidence (P-value > .20) to conclude that there is a difference in resting blood concentration of HBE between joggers and fitness program entrants.

7.93: Let 1 denote Vermilion River and let 2 denote Black River.

H0: The populations from which the two samples were drawn have the same distribution of tree species per plot.

HA: Biodiversity is greater along the Vermilion River than along the Black River.

K1 = 37, K2 = 80, so Us = 80; the data deviate from H0 in the direction specified by HA. With n = n’ = 13 and a directional alternative, the .10 entry in Table 6 is 79 and the .05 entry is 84. Thus, the P-value is bracketed between .05 and .10, so we reject H0. There is sufficient evidence (.05 < P-value < .10) to conclude that biodiversity is greater along the Vermilion River than along the Black River.