chapter 7 comparison of two independent samples · · 2006-04-10using df = 1000 (the nearest...

CHAPTER 7 Comparison of Two Independent Samples

•7.1 We first find the standard error of each mean.

SE1 = SEy-1 =

s1n1

= 4.3

6 = 1.755.

SE2 = SEy-2 =

s2n2

= 5.712 = 1.645.

SE(y-1 - y-2) = SE12 + SE22 = 1.7552 + 1.6452 = 2.41.

7.2 SE1 = 44.2

10 = 13.977; SE2 = 28.7

10 = 9.076.

13.9772 + 9.0762 = 16.7.

7.3 SE1 = 525 = 1; SE2 =

629 = 1.114.

12 + 1.1142 = 1.50.

7.4 SE1 = 6.5

5 = 2.907; SE2 = 8.4

7 = 3.175.

2.9072 + 3.1752 = 4.30.

7.5 SE1 = 6.510 = 2.055; SE2 =

8.414 = 2.245.

2.0552 + 2.2452 = 3.04.

7.6 3.72 + 4.62 = 5.90.

•7.7 SE(y-1 - y-2) = SE12 + SE22 = .52 + .72 = .86.

7.8 SE1 = .400

9 = .133; SE2 = .220

6 = .090.

.1332 + .0902 = .16.

7.9 5.52 + 8.62 = 10.2.

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•7.10 Let 1 denote males and let 2 denote females. y- 1 = 45.8; SE1 = 2.8/ 489 = .127. y- 2 = 40.6; SE2 = 2.9/ 469 = .134.

The standard error of the difference is SE(y-1 - y-2) = .1272 + .1342 = .185.

The critical value t.025 is determined from Student's t distribution with df = 950. Using df = 1000 (the nearest value given in Table 4), we find that t(1000).025 = 1.962.

The 95% confidence interval is y- 1 - y- 2 ± t.025SE(y-1 - y-2)

(45.8 - 40.6) ± (1.962)(.185).

So the confidence interval is (4.84,5.56) or 4.84 < µ1 - µ2 < 5.56.

7.11 Let 1 denote dark and let 2 denote photoperiod.

SE(y-1 - y-2) = 132

4 + 132

4 = 9.192.

(a) (92 - 115) ± (2.447)(9.192) (df = 6)

(-45.5,-.5) or -45.5 < µ1 - µ2 < -.5 nmol/gm.

( b) (92 - 115) ± (1.943)(9.192) (df = 6)

(-40.9,-5.1) or -40.9 < µ1 - µ2 < -5.1 nmol/gm.

7.12 (a) Let 1 denote biofeedback and let 2 denote control. SE(y-1 - y-2) = 1.342 +1.302 = 1.867.

(13.8 - 4.0) ± (1.977)(1.867) (using df = 140)

(6.1,13.5) or 6.1 < µ1 - µ2 < 13.5 mm Hg.

(b) We are 95% confident that the population mean reduction in systolic blood pressure for those who receive training for eight weeks (µ1) is larger than that for others (µ2) by an amount that might be as small as 6.1 mm Hg or as large as 13.5 mm Hg.

7.13 No. The confidence interval found in Exercise 7.11 is valid even if the distributions are not normal,

because the sample sizes are large.

•7.14 (a) Let 1 denote antibiotic and let 2 denote control.

SE(y-1 - y-2) =102

10 + 82

10 = 4.050.

(25 - 23) ± (1.740)(4.050) (using df = 17)

(-5.0,9.0) or -5 < µ1 - µ2 < 9 sec.

(b) We are 90% confident that the population mean prothrombin time for rats treated with an antibiotic (µ1) is smaller than that for control rats (µ2) by an amount that might be as much as 5 seconds or is larger than that for control rats (µ2) by an amount that might be as large as 9 seconds.

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7.15 Let 1 denote control and let 2 denote Pargyline.

SE(y-1 - y-2) = 5.42

900 + 11.72

905 = .4286.

(a) (14.9 - 46.5) ± (1.96)(.4286) (df = ∞)

(-32.4,-30.8) or -32.4 < µ1 - µ2 < -30.8 mg.

(b) (14.9 - 46.5) ± (2.576)(.4286) (df = ∞)

(-32.7,-30.5) or -32.7 < µ1 - µ2 < -30.5 mg.

7.16 (a) Let 1 denote successful and let 2 denote unsuccessful.

SE(y-1 - y-2) =.2832

22 + .2622

17 = .08763.

(8.498 - 8.440) ± (2.021)(.08763) (using df = 40)

(-.12,.24) or -.12 < µ1 - µ2 < .24 mm.

(b) We are 95% confident that the population mean head width of all females who mate successfully (µ1) is smaller than that for rejected females (µ2) by an amount that might be as much as .12 mm or is larger than that for rejected females (µ2) by an amount that might be as large as .24 mm.

7.17 We are 97.5% confident that the population mean drop in systolic blood pressure of adults placed on a diet

rich in fruits and vegetables for eight weeks (µ1) is larger than that for adults placed on a standard diet (µ2) by an amount that might be as small as .9 mm Hg or as large as 4.7 mm Hg.

7.18 We are 97.5% confident that the population mean drop in diastolic blood pressure of adults placed on a

diet rich in fruits and vegetables for eight weeks (µ1) is smaller than that for adults placed on a standard diet (µ2) by an amount that might be as much as .3 mm Hg or is larger than that for adults placed on a standard diet (µ2) by an amount that might be as much as 2.4 mm Hg.

7.19 Let 1 denote caffeine and let 2 denote decaf.

SE(y-1 - y-2) = 3.72 +3.42 = 5.02.

(7.3 – 5.9) ± (1.740)(5.02) (using df = 17)

(-7.33,10.13) or -7.33 < µ1 - µ2 < 10.13. 7.20 Normal probability plots support the normality condition:

-0.0

7.5

15.0

22.5

-0.75 0.75

nscores

caffeine

-10

0

10

20

-1 0 1

nscores

decaf

7.21 Let 1 denote red and let 2 denote green.

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SE(y-1 - y-2) = .362 + .362 = .509.

(8.36 – 8.94) ± (2.021)(.509) (using df = 40)

(-1.61,0.45) or -1.61 < µ1 - µ2 < 0.45.

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.

•7.23 (a) The observed t statistic is

ts = y-1 - y-2

SE(y-1 - y-2) =

735 - 85438 = -3.13.

From Table 4 with df = 4, we find the critical values t.02 = 2.999 and t.01 = 3.747. Because ts is between t.01 and t.02, the P-value must be between .02 and .04. Thus, the P-value is bracketed as .02 < P < .04.

(b) The observed t statistic is

ts = y-1 - y-2

SE(y-1 - y-2) =

5.3 - 5.0.24 = 1.25.

From Table 4 with df = 12, we find the critical values t.20 = .873 and t.10 = 1.356. Because ts is between t.10 and t.20, the P-value must be between .20 and .40. Thus, the P-value is bracketed as .20 < P < .40.

(c) The observed t statistic is

ts = y-1 - y-2

SE(y-1 - y-2) =

36 - 301.3 = 4.62.

From Table 4 with df = 30, we find the critical value t.0005 = 3.646. Because ts is greater than t.0005, the P-value must be less than .001. Thus, the P-value is bracketed as P < .001.

7.24 (a) ts = (100.2 - 106.8)/5.7 = -1.16.

t(10).20 = .879; t(10).10 = 1.372. Thus, .20 < P < .40.

(b) ts = (49.8 - 44.3)/1.9 = 2.89.

t(13).01 = 2.650; t(13).005 = 3.012. Thus, .01 .05, which means that the P-value is greater than α. Thus, we do not reject H0.

(c) Table 4 gives t(19).005 = 2.861 and t(19).0005 = 3.883, so .001 α, we do not reject H0.

7.26 (a) Because P > α, we do not reject H0. (b) Because P < α, we reject H0. (c) t(5).04 = 2.191 and t(5).03 = 2.422, so .06 < P < .08. Because P < α, we reject H0.

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(d) t(16).04 = 1.869 and t(16).03 = 2.024, so .06 α, we do not reject H0.

Remark concerning tests of hypotheses The answer to a hypothesis testing exercise includes verbal statements of the hypotheses and a verbal statement of the conclusion from the test. In phrasing these statements, we have tried to capture the essence of the biological question being addressed; nevertheless the statements are necessarily oversimplified and they gloss over many issues that in reality might be quite important. For instance, the hypotheses and conclusion may refer to a causal connection between treatment and response; in reality the validity of such a causal interpretation usually depends on a number of factors related to the design of the investigation (such as unbiased allocation of animals to treatment groups) and to the specific experimental procedures (such as the accuracy of assays or measurement techniques). In short, the student should be aware that the verbal statements are intended to clarify the statistical concepts; their biological content may be open to question.

7.27 H0: µHH = µSH

HA: µHH ≠ µSH

SE(y-1 - y-2) = 17.82

33 + 19.12

51 = 4.09; ts = 18.3 - 13.9

4.09 = 1.07.

df = 71.9, using formula (7.1), or df = 32 using the conservative choice of min{n1-1,n2-1}, or df = 82 using n1 + n2 - 2. In any case, .20 α and we do not reject

H0.

7.28 y- 1 - y- 2 = 3.49 - 3.05 = .44.

(a) SE(y-1 - y-2) =.42

5 + .42

5 = .2530

ts = .44/.2530 = 1.74.

Table 4 gives t(8).10 = 1.397 and t(8).05 = 1.860, so .10 < P < .20.

(b) SE(y-1 - y-2) =.42

10 + .42

10 = .1789.

ts = .44/.1789 = 2.46.


(c) SE(y-1 - y-2) =.42

15 + .42

15 = .1461.

ts = .44/.1461 = 3.01.


•7.29 (a) The null and alternative hypotheses are

H0: µ1 = µ2 HA: µ1 ≠ µ2

where 1 denotes heart disease and 2 denotes control. These hypotheses may be stated as

H0: Mean serotonin concentration is the same in heart patients and in controls HA: Mean serotonin concentration is not the same in heart patients and in controls

The test statistic is

101

ts = y-1 - y-2

SE(y-1 - y-2) =

3840 - 53101064 = -1.38.

From Table 4 with df = 14, we find the critical values t.10 = 1.345 and t.05 = 1.761. Thus, the P-value is bracketed as .10 < P < .20.

Since the P-value is greater than α (.05), H0 is not rejected.

(b) There is insufficient evidence (.10 < P < .20) to conclude that serotonin levels are different in heart patients than in controls.

(c) SE(y-1 - y-2) = 8502 + 6402 = 1064.

7.30 (a) H0: mean tibia length does not depend on gender (µ1 = µ2)

HA: mean tibia length depends on gender (µ1 ≠ µ2)

SE(y-1 - y-2) = 2.872

60 + 3.522

50 = .62056.

ts = (78.42 - 80.44)/.62056 = -3.26. df = n1 + n2 - 2 = 108 ≈ 100. (Formula (7.1) gives df = 94.3.) Table 4 gives t.005 = 2.626 and t.0005 = 3.390; thus .001 < P < .01, so we reject H0.

(b) There is sufficient evidence (.001 < P < .01) to conclude that mean tibia length is larger in females than in males.

(c) Judging from the means and SDs, the two distributions overlap substantially, so tibia length would be a poor predictor of sex.

(d) H0 and HA are as is part (a).

SE(y-1 - y-2) = 2.872

6 + 3.522

5 = 1.962

ts = (78.42 - 80.44)/1.962 = -1.03. df = n1 + n2 - 2 = 9. (Formula (7.1) gives df = 7.8.) Table 4 gives t.20 = .883 and t.10 = 1.383; thus .20 < P < .40 and we do not reject H0.

7.31 (a) H0: mean thymus weight is the same at 14 and 15 days (µ1 = µ2)

HA: mean thymus weight is not the same at 14 and 15 days (µ1 ≠ µ2)

SE(y-1 - y-2) = 8.732

5 + 7.192

5 = 5.06

ts = (31.72 - 29.22)/5.06 = .49. df = n1 + n2 - 2 = 8. (Formula (7.1) gives df = 7.7.) Table 4 gives t.20 = .889; thus P > .40, so we do not reject H0. There is insufficient evidence (P > .40) to conclude that mean thymus weight is different at 14 and 15 days.

(b) According to the P-value found in part (a), the fact that y- 1 is greater than y- 2 could easily be attributed to chance.

[Remark: A student has commented that it would not be surprising if µ1 were actually less than µ2, because

in certain cases the thymus gland would be expected to shrink during embryonic development.]


H0: µ1 = µ2 HA: µ1 ≠ µ2

102

where 1 denotes flooded and 2 denotes control. These hypotheses may be stated as

H0: Flooding has no effect on ATP HA: Flooding has some effect on ATP

The standard error of the difference is

SE(y-1 - y-2) =.1842

4 + .2412

4 = .1516.


ts = y-1 - y-2

SE(y-1 - y-2) =

1.190 - 1.785.1516 = -3.92.

From Table 4 with df = n1 + n2 - 2 = 6 (Formula (7.1) yields df = 5.6), we find the critical values t.005 = 3.707 and t.0005 = 5.959. Thus, the P-value is bracketed as .001 < P < .01. Since the P-value is less than α (.05), we reject H0.

(b) There is sufficient evidence (.001 < P < .01) to conclude that flooding tends to lower ATP in birch seedlings.

7.33 (a) H0: Albumin and polygelatin are equally effective as plasma expanders (µ1 = µ2) HA: Albumin and polygelatin are not equally effective as plasma expanders (µ1 ≠ µ2)

SE(y-1 - y-2) = 602 + 302 = 67.08.

ts = (490 - 240)/67.08 = 3.73.

df = n1 + n2 - 2 = 37 ≈ 40 and t(40).0005 = 3.551. Thus, P < .001, so we reject H0.

(b) There is sufficient evidence (P < .001) to conclude that albumin is more effective than polygelatin as a plasma expander.

7.34 H0: Mean fall in cholesterol is the same on both diets (µ1 = µ2) HA: Mean fall in cholesterol is not the same on both diets (µ1 ≠ µ2)

SE(y-1 - y-2) =31.12

10 + 29.42

10 = 13.53.

ts = (53.6 - 55.5)/13.53 = -.14.

df = n1 + n2 - 2 = 18 and t(18).20 = .862. Thus, P > .40, so we do not reject H0. There is insufficient evidence (P > .40) to conclude that the two diets differ in their effects on cholesterol.

7.35 (a) True. We would reject H0 because the P-value is less than α.

(b) True. We would reject H0 because the P-value is less than α. (c) True. This follows directly from the definition of a P-value.

7.36 (a) True. We would reject H0 because the P-value is less than α.

(b) False. We do not reject H0 because the P-value is greater than α.

(c) False. The P-value is the probability, under H0, of getting a result as extreme as, or more extreme than, the result that was actually observed.

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2 2

7.37 (a) H0: mean number of colonies is the same for control and soap (µ1 = µ2) HA: mean number of colonies is not the same for control and soap (µ1 ≠ µ2)

SE(y-1 - y-2) = 5.5 +8.6 = 10.21

ts = (41.8 – 32.4)/10.21 = .92. Using Table 4 with df=10 we have t.20 = .879 and t.10 = 1.372. Thus .20 < P < .40, so we do not reject H0.

(b) There is insufficient evidence (.20 < P < .40) to conclude that the mean number of colonies differs for control and soap.

7.38 (a) H0: mean height is the same for control and fertilized plants (µ1 = µ2) HA: mean height is not the same for control and fertilized plants (µ1 ≠ µ2)

SE(y-1 - y-2) = .6528

2

+.7228

2

= .183

ts = (2.58 – 2.04)/.183 = 2.95. Using Table 4 with df=50 we have t.005 = 2.678 and t.0005 = 3.496. Thus .001 < P < .01, so we reject H0.

(b) There is strong evidence (.001 < P < .01) to conclude that the mean height of fertilized radish sprouts is less than that of controls.

7.39 - 7.41 See Section III of this Manual.

•7.42 If we reject H0 (i.e., if the drug is approved) then we eliminate the possibility of a Type II error. (But by rejecting H0 we may have made a Type I error.)

7.43 A type II error may have been made.

•7.44 Yes; because zero is outside of the confidence interval, we know that the P-value is less than .05, so we reject the hypothesis that µ1 - µ2 = 0.

7.45 Yes; because zero is outside of the confidence interval, we know that the P-value is less than .05, so the P-value is less than .10. Thus, we reject the hypothesis that µ1 - µ2 = 0.

•7.46 (a) The observed t statistic is

ts = y-1 - y-2

SE(y-1 - y-2) =

10.8 - 10.5.23 = 1.30.

To check the directionality of the data, we note that y- 1 > y- 2. Thus, the data do deviate from H0 in the direction (µ1 > µ2) specified by HA, and therefore the one-tailed P-value is the area under the t curve beyond 1.30. From Table 4 with df = 18, we find the critical values t.20 = .862 and t.10 = 1.330. Because ts is between t.10 and t.20, the one-tailed P-value must be between .10 and .20. Thus, the P-value is bracketed as .10 < P < .20.

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(b) The observed t statistic is

ts = y-1 - y-2

SE(y-1 - y-2) =

750 - 73011 = 1.82.

To check the directionality of the data, we note that y- 1 > y- 2. Thus, the data do deviate from H0 in the direction (µ1 > µ2) specified by HA, and therefore the one-tailed P-value is the area under the t curve beyond 1.82. From Table 4 with df = 140 (the closest value to 180), we find the critical values t.04 = 1.763 and t.03 = 1.896. Because ts is between t.03 and t.04, the one-tailed P-value must be between .03 and .04. Thus, the P-value is bracketed as .03 .20.

(b) ts = 560 - 500

45 = 1.33.

With df = 8, Table 4 gives t.20 = .889 and t.10 = 1.397. Thus, .10 .50.

•7.48 (a) Yes. ts is positive, as predicted by HA. Thus, the P-value is the area under the t curve beyond 3.75. With df = 19, Table 4 gives t.005 = 2.861 and t.0005 = 3.883. Thus, .0005 < P < .005. Since P < α, we reject H0.

(b) Yes. ts is positive, as predicted by HA. Thus, the P-value is the area under the t curve beyond 2.6. With df = 5, Table 4 gives t.025 = 2.571 and t.02 = 2.757. Thus, .02 < P < .025. Since P < α, we reject H0.

(c) Yes. ts is positive, as predicted by HA. Thus, the P-value is the area under the t curve beyond 2.1. With df = 7, Table 4 gives t.04 = 2.046 and t.03 = 2.241. Thus, .03 < P < .04. Since P < α, we reject H0.

(d) No. ts is positive, as predicted by HA. Thus, the P-value is the area under the t curve beyond 1.8. With df = 7, Table 4 gives t.10 = 1.415 and t.05 = 1.895. Thus, .05 α, we do not reject H0.

7.49 (a) No. With df = 23, Table 4 gives t.10 = 1.319 and t.05 = 1.714. Thus, .05 α, we do not reject H0.

(b) Yes. With df = 5, Table 4 gives t.04 = 2.191 and t.03 = 2.422. Thus, .03 0, the data do not deviate from H0 in the direction specified by HA. Thus, P > .50 and we do not reject H0.

(d) Yes. With df = 27, Table 4 gives t.005 = 2.771 and t.0005 = 3.690. Thus, .0005 < P < .005. Since P < α, we reject H0.

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7.50 Let 1 denote infected and 2 denote noninfected.

H0: Malaria does not affect red cell count (µ1 = µ2) HA: Malaria reduces red cell count (µ1 < µ2)

We note that y- 1 > y- 2, so the data do not deviate from H0 in the direction specified by HA. Thus, P > .50.

(a) H0 is not rejected. There is no evidence that malaria reduces red cell count in this population.

(b) Same as part (a).

[Note: If HA were reversed (µ1 > µ2), then H0 would be rejected at α = .10. Thus, this exercise illustrates the importance of the directionality check.]

7.51 Let 1 denote experimental (to be hypnotized) and 2 denote control.

SE(y-1 - y-2) =.6212

8 + .6522

8 = .3183.

ts = (6.169 - 5.291)/.3183 = 2.76.

With df = n1 + n2 - 2 = 14 (Formula (7.1) yields df = 13.97), Table 4 gives t.01 = 2.624 and t.005 = 2.977.

(a) H0: Mean ventilation is the same in the "to be hypnotized" condition and in the "control" condition (µ1 = µ2) HA: Mean ventilation is different in the "to be hypnotized" condition than in the "control" condition (µ1 ≠ µ2)

H0 is rejected. There is sufficient evidence (.01 < P < .02) to conclude that mean ventilation is higher in the "to be hypnotized" condition than in the "control" condition.

(b) H0: Mean ventilation is the same in the "to be hypnotized" condition and in the "control" condition (µ1 = µ2) HA: Mean ventilation is higher in the "to be hypnotized" condition than in the "control" condition (µ1 > µ2)

H0 is rejected. There is sufficient evidence (.005 < P < .01) to conclude that mean ventilation is higher in the "to be hypnotized" condition than in the "control" condition.

(c) The nondirectional alternative (part (a)) is more appropriate. According to the narrative, the researchers formulated the directional alternative in part (b) after they had seen the data. Thus, it would not be legitimate for them (or us) to use a directional alternative.


H0: Extra nitrogen does not enhance plant growth (µ1 = µ2) HA: Extra nitrogen does enhance plant growth (µ1 < µ2)

SE(y-1 - y-2) =.542

5 + .672

5 = .3848.

ts = (3.62 - 4.17)/.3848 = -1.43.

With df = n1 + n2 - 2 = 8 (Formula (7.1) yields df = 7.7), Table 4 gives t.10 = 1.397 and t.05 = 1.860.

H0 is not rejected. There is insufficient evidence (.05 < P < .10) to conclude that extra nitrogen enhances plant growth under these conditions.

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•7.53 The null and alternative hypotheses are

H0: µ1 = µ2 HA: µ1 < µ2

where 1 denotes wounded and 2 denotes control. These hypotheses may be stated as

H0: Wounding the plant has no effect on larval growth HA: Wounding the plant tends to diminish larval growth

To check the directionality of the data, we note that y- 1 < y- 2. Thus, the data do deviate from H0 in the direction (µ1 < µ2) specified by HA. We proceed to calculate the test statistic. The standard error of the difference is

SE(y-1 - y-2) =9.022

16 + 11.142

18 = 3.46.


ts = y-1 - y-2

SE(y-1 - y-2) =

28.66 - 37.963.46 = -2.69.

From Table 4 with df = 16 + 18 -2 = 32 ≈ 30 (Formula (7.1) yields df = 31.8), we find the critical values t.01 = 2.457 and t.005 = 2.750. Thus, the P-value is bracketed as .005 µ2

where 1 denotes drug and 2 denotes placebo. These hypotheses may be stated as

H0: The drug is not effective HA: The drug is effective To check the directionality of the data, we note that y- 1 > y- 2. Thus, the data do deviate from H0 in the direction (µ1 > µ2) specified by HA. We proceed to calculate the test statistic. The standard error of the difference is

SE(y-1 - y-2) =12.052

25 + 13.782

25 = 3.66.


ts = y-1 - y-2

SE(y-1 - y-2) =

31.96 - 25.323.66 = 1.81.

From Table 4 with df = 25 + 25 -2 = 48 ≈ 50 (Formula (7.1) yields df = 47.2), we find the critical values t.04 = 1.787 and t.03 = 1.924. Thus, the P-value is bracketed as

.03 < P < .04. Since the P-value is less than α (.05), we reject H0. There is sufficient evidence (.03 < P < .04) to conclude that the drug is effective at increasing pain relief.

(b) The only change in the calculations from part (a) would be that the one-tailed area would be doubled if the alternative were nondirectional. Thus, the p-value would be between .06 and .08 and at α = .05 we would not reject H0.

107

7.55 .03 < P-value < .05. Thus, we would reject H0 and conclude that ancy does, indeed, inhibit growth.

7.56 Let 1 denote 250 meters and 2 denote 800 meters.

H0: Distance from the reef does not affect settler density (µ1 = µ2) HA: Settler density decreases as distance from the reef increases (µ1 > µ2)

SE(y-1 - y-2) =.5142

48+

.4132

48= .095.

ts = (.818 - .628)/.095 = 2.

With df =80, Table 4 gives t.025 = 1.990 and t.02 = 2.088. Thus, .02 < P < .025 and we reject H0. There is statisitcally significant evidence (.02 < P < .025) to conclude that settler density decreases as distance from the reef increases.

7.57 The proponents are confused. They are speaking as if it is known that µ1 - µ2 = 4 lb/acre, whereas the field trial indicates only that y- 1 - y- 2 = 4 lb/acre. That statistician's data analysis indicates that the trial gives only weak information about µ1 - µ2; in fact, the results do not even show whether µ1 - µ2 is positive, let alone that it is equal to 4 lb/acre.

7.58 The lack of a statistically significant difference in therapeutic responses does not show that the two medications are equally effective. (Such evidence could be obtained from either a confidence interval or an analysis of the power of the test.)

7.59 Let 1 denote male and 2 denote female.

SE(y-1 - y-2) = .622 + .532 = .8157.

(137.21 - 137.18) ± (1.977)(.8157) (using df = 140)

(-1.6,1.6) or -1.6 < µ1 - µ2 < 1.6 beats per minute.

We can be 95% confident that the mean difference does not exceed 1.6 beats per minute, which is small and unimportant (in comparison with, for example, ordinary fluctuations in heart rate from one minute to the next.)

•7.60 The mean difference in concentration of coumaric acid is µ1 - µ2, where 1 denotes dark and 2 denotes photoperiod. We construct a 95% confidence interval for µ1 - µ2.

SE(y-1 - y-2) =212

4 + 272

4 = 17.103.

The critical value t.025 is found from Student's t distribution with df = n1 + n2 - 2 = 6. (Formula (7.1) gives df = 5.7.) From Table 4, we find t(6).025 = 2.447. The 95% confidence interval is y- 1 - y- 2 ± t.025SE(y-1 - y-2)

(106 - 102) ± (2.447)(17.103)

(-37.9,45.9) or -37.9 < µ1 - µ2 < 45.9 nmol/gm.

The difference could be larger than 20 nmol/gm or much smaller, so the data do not indicate whether the difference is "important."

108

7.61 SE(y-1 - y-2) =212

40+

272

40= 5.408.

(106 - 102) ± (1.994)(5.408)

(-6.8,14.8) or -6.8 < µ1 - µ2 < 14.8 nmol/gm.

The difference is less than 20 nmol/gm, so the data do indicate that the difference is "important."

•7.62 The mean difference in serum concentration of uric acid is µ1 - µ2, where 1 denotes men and 2 denotes women. We construct a 95% confidence interval for µ1 - µ2.

SE(y-1 - y-2) =.0582

530 + .0512

420 = .00354.

The critical value t.025 is found from Student's t distribution with df = n1 + n2 - 2 = 948 ≈ 1000. (Formula (7.1) gives df = 937.8.) From Table 4, we find t(1000).025 = 1.962. The 95% confidence interval is y- 1 - y- 2 ± t.025SE(y-1 - y-2)

(.354 - .263) ± (1.962)(.00354)

(.0841,.0979) or .0841 < µ1 - µ2 < .0979 mmol/l.

All values in the confidence interval are greater than .08 mmol/l. Therefore, according to the confidence interval the data indicate that the difference is "clinically important."

7.63 SE(y-1 - y-2) =.0582

53 + .0512

42 = .0112.

(.354 - .263) ± (1.984)(.0112) (using df = 100)

(.069,.113) or .069 < µ1 - µ2 < .113 mmol/l.

The difference could be greater than or less than .08 mmol/l, so the data do not indicate whether the difference is "clinically important."

•7.64 From the preliminary data, we obtain .3 cm as a guess of σ. (a) If the true difference is .25 cm, then the effect size is

|µ1 - µ2|

σ = .25.3 = .83.

We consult Table 5 for a two-tailed test at α = .05 and an effect size of .83 ≈ .85; to achieve power .80, Table 5 recommends n = 23.

(b) If the true difference is .5 cm, then the effect size is

|µ1 - µ2|

σ = .5.3 = 1.67.

We consult Table 5 for a two-tailed test at α = .05 and an effect size of 1.67 ≈ 1.7; to achieve power .95, Table 5 recommends n = 11.

7.65 |µ1 - µ2|/σ = 44.4/69.6 = .64.

(a) Table 5 gives n = 39.

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(b) Table 5 gives n = 30.

7.66 |µ1 - µ2|/σ = 2/.8 = 2.5.


(b) The required conditions are that the sampled populations are normal with equal standard deviations. The condition of normality can be checked from the pilot data.

(c) Table 5 gives n = 7.

•7.67 We need to find n to achieve a power of .9. The effect size is

|µ1 - µ2|

σ = 81 - 75

11 = .55.

We consult Table 5. (a) For a two-tailed test at α = .05, Table 5 gives n = 71. (b) For a two-tailed test at α = .01, Table 5 gives n = 101. (c) For a one-tailed test at α = .05, Table 5 gives n = 58.

7.68 (a) |µ1 - µ2|/σ = 2/2.5 = .8; Table 5 gives n = 26. (b) |µ1 - µ2|/σ = 2/1.25 = 1.6; Table 5 gives n = 8. Yes, the measure would be cost-effective, because the required number of rats would be reduced by more than half. (In fact, the modified experiment would cost only 62% as much, because (2)(8)/26 = .62.)

•7.69 The effect size is

|µ1 - µ2|

σ = 410 = .4.

From Table 5 we find that, for a one-tailed test with n = 35 at significance level α = .05, the power is .50 if the effect size is .4. Thus, the probability that Jones will reject H0 is equal to .50.

7.70 |µ1 - µ2|/σ = .5 = .8.


(b) Table 5 gives n = 70.

7.71 (a)

(b)

110

7.72 n = 13

7.73 |µ1 - µ2|/σ = 1.2

7.74 (a) Both of the distributions appear to be slightly skewed to the right.

(b) Because n1 and n2 are quite large, Student's t method is valid even if the populations are nonnormal.

7.75 The sample SDs are:

Heart disease: s1 = (850) 8 = 2404

Controls: s2 = (640) 12 = 2217

The fact that the mean (3840) for the heart disease patients is only about 1.6 SDs, together with the fact that the observed variable cannot be negative, casts doubt on the normality condition for the population. Because the sample sizes are rather small, such a departure from normality can lead to poor performance by the t test.

7.76 (a) False; the P-value is the probability of data as unusual as those obtained if H0 is true.

(b) True; reject H0 since the P-value is less than α.

(c) False. We should reject H0, but don’t know the chance that we would reject H0 in repeated experiments – we do not know if H0 is true.

(d) True; this is the interpretation of a P-value.

•7.77 We consult Table 6 with n = 7 and n' = 5.

(a) Us = 26. From Table 6, the smallest critical value is 27, which corresponds to a nondirectional P-value of .20. Since Us < 27, it follows that P > .20.

(b) Us = 30. From Table 6, the critical value 30 is under the .05 heading for a nondirectional alternative. The P-value is between this heading and the next one to the right. Thus, .02 < P .05.

(c) Us = 35. From Table 6, the largest critical value is 34, which is under the .01 heading for a nondirectional alternative. There is no entry under the .002 heading, which means that the P-value is greater than .002. It follows that .002 < P < .01.

7.78 n = 8, n' = 4.

(a) .10 < P < .20.

(b) .002 < P < .01.

111

(c) The .002 entry is blank, so .002 < P < .01.


H0: Toluene has no effect on dopamine in rat striatum HA: Toluene has some effect on dopamine in rat striatum

Let 1 denote toluene and let 2 denote control. The ordered arrays of observations are as follows:

Y1

:

Y2:

1911 2314 34202464 2781 2803

1820 18431397 1803 25391990

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, or fourth Y2; there is one Y1 less than the fifth Y2; and there 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

.02 < P < .05. At significance level α = .05, we reject H0, since P < .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 (.02 < P < .05) to conclude that toluene increases dopamine in rat striatum.

(b) When conducting a nondirectional 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, .01 < P < .025. We reject H0 and conclude that there is sufficient evidence (.01 < P < .025) to conclude that toluene increases dopamine in rat striatum.


(a) H0: Ventilation is not differently affected by the "to be hypnotized" and the "control" conditions HA: Ventilation is differently affected by the "to be hypnotized" and the "control" conditions

K1 = 53, K2 = 11, Us = 53. With n = 8 and n' = 8, Table 6 gives .02 < P < .05. H0 is rejected. There is sufficient evidence (.02 < P < .05) to conclude that ventilation rate tends to be higher under the "to be hypnotized" condition than under the "control" condition.

7.81 (a) Us = 9. With n = n' = 3, Us = 9 is under the .10 heading and is the largest entry listed. Thus, .05 < P < .10.

(b) Us = 16. With n = n' = 4, Us = 16 is under the .05 heading and is the largest entry listed. Thus,

112

.02 < P < .05.

(c) Us = 25. With n = n' = 5, Us = 25 is under the .01 heading and is the largest entry listed. Thus, .002 < P < .01.

7.82 (a) H0: There is no sex difference in preening behavior HA: There is a sex difference in preening behavior

For n = n' = 15, the largest critical value is 189, which is under the .001 heading for a nondirectional alternative. It follows that P < .001, so H0 is rejected. There is sufficient evidence (P < .001) to conclude that females tend to preen longer than males.

(b) H0: There is no sex difference in preening behavior (µ1 = µ2) HA: There is a sex difference in preening behavior (µ1 ≠ µ2)

ts = (2.127 - 4.093)/.7933 = -2.48. With df = n1 + n2 - 2 = 28, Table 4 gives t.01 = 2.467 and t.005 = 2.763, so that .01 < P < .02. Formula (7.1) yields df = 15.1 and the conservative approach of df = min{n1 - 1,n2 - 1} gives df = 14. For either of these df values we get .02 .01. There is sufficient evidence to conclude that there is a sex difference in preening behavior.

(c) Both tests require independent, random samples. The condition required for the t test but not for the Wilcoxon-Mann-Whitney test is that the population distributions are normal. The frequency distribution for the females is highly skewed, due to the two large observations of 10.7 and 11.7. This casts doubt on the normality condition.

(d) K1 = 0 + 0 + 0 + 0 + 0 + .5 + 1 + 1.5 + 1.5 + 2 + 2 + 3.5 + 5 + 8.5 + 10 = 35.5 K2 = 5.5 + 8 + 3(11.5) + 3(13) + 13.5 + 14 + 5(15) = 189.5

where 1 denotes male and 2 denotes female.

7.83 (a) Let 1 denote singly housed and let 2 denote group-housed.

H0: There is no difference in benzo(a)pyrene concentrations between singly housed and group-housed mice. HA: Benzo(a)pyrene concentrations tend to be higher in group-housed mice than in singly housed mice

K1 = 0, K2 = 25, Us = 25, and the shift in the data is in the direction predicted by HA. With n = n' = 5, Us = 25 is under the .005 heading for a directional alternative and is the largest entry listed. Thus, .001 < P < .005 and H0 is rejected. There is sufficient evidence (.001 < P < .005) to conclude that benzo(a)pyrene concentrations tend to be higher in group-housed mice than in singly housed mice.

(b) A directional alternative is valid in this case because the researchers were investigating the hypothesis that licking or biting other mice leads to increase benzo(a)pyrene concentration. If access to other mice affects benzo(a)pyrene concentration, the effect would be to increase the concentration; a decrease in concentration is not plausible.

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 fitness program entrants HA: There is a difference in resting blood concentration of HBE between joggers 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 alternative and is the smallest entry listed. Thus, P > .20 and H0 is not rejected. There is insufficient evidence (P > .20) to conclude that there is a difference in resting blood concentration of HBE

113

between joggers fitness program entrants.

7.85 (a) SE1 = 9.6/ 12 = 2.771; SE2 = 10.2/ 13 = 2.829. 2.7712 + 2.8292 = 3.96.

(b) SE1 = 2.7/ 22 = .576; SE2 = 1.9/ 19 = .436. .5762 + .4362 = .722.

(c) SE1 = 1.2/ 5 = .537; SE2 = 1.4/ 7 = .529. .5372 + .5292 = .754.


H0: Mean platelet calcium is the same in people with high blood pressure as in people with normal blood pressure (µ1 = µ2) HA: Mean platelet calcium is different in people with high blood pressure than in people with normal blood pressure (µ1 ≠ µ2)


SE(y-1 - y-2) = 16.12

38+

31.72

45 = 5.399.


ts = y-1 - y-2

SE(y-1 - y-2) =

168.2 - 107.95.399 = 11.2.

From Table 4 with df = 45 + 38 -2 = 81 ≈ 80, we find the critical value t.0005 = 3.416. The tail area is doubled for the nondirectional test. Thus, the P-value is bracketed as P < .001. (Formula (7.1) yields df = 67.5, but the P-value is still bracketed as P < .001.)

Since the P-value is less than α (.01), we reject H0. There is sufficient evidence (P < .001) to conclude that mean platelet calcium is higher in people with high blood pressure than in people with normal blood pressure.

•7.87 The mean difference in blood pressure is µ1 - µ2, where 1 denotes high blood pressure and 2 denotes

normal blood pressure.


SE(y-1 - y-2) =31.72

45 + 16.12

38 = 5.399.

The critical value t.025 is found from Student's t distribution with df given by Formula (7.1) as df = 67.5 ≈ 70. Table 4 gives t(70).025 = 1.994. The 95% confidence interval is

y- 1 - y- 2 ± t.025SE(y-1 - y-2)

(168.2 - 107.9) ± (1.994)(5.399).

So the confidence interval is (49.5,71.1) or 49.5 < µ1 - µ2 < 71.1 nM. Alternatively, we could use df = 45 + 38 -2 = 81 ≈ 80, in which case the critical value is t(80).025 = 1.990.

This gives an interval of

(168.2 - 107.9) ± (1.990)(5.399).

114

So the confidence interval is (49.6,71.0) or 49.6 < µ1 - µ2 < 71.0 nM.

7.88 No; the t test is valid because the sample sizes are rather large. 7.89 (a) H0: Mechanical milking does not produce different cell count than manual milking (µ1 = µ2)

HA: Mechanical milking produces higher cell count than manual milking (µ1 > µ2)

ts = (1215.6 - 219.0)/427.54 = 2.33. With df = 18, Table 4 gives t.02 = 2.214 and t.01 = 2.552. Formula (7.1) yields df = 9.2 ≈ 9; with df = 9, Table 4 gives t.025 = 2.262 and t.02 = 2.398. Using either df value, P < .05 and H0 is rejected. There is sufficient evidence to conclude that mechanical milking produces higher cell count than manual milking. (The data support the investigator's suspicion.)

(b) H0: Mechanical milking does not produce different cell count than manual milking HA: Mechanical milking produces higher cell count than manual milking

Us = 69. The shift in the data is in the direction predicted by HA. With n = n' = 10, the entry under .10 for a directional alternative is 68 and the entry under .05 for a directional alternative is 73. Thus, we do not reject H0. There is insufficient evidence (.05 < P < .10) to conclude that mechanical milking produces higher cell count than manual milking. (The data do not support the investigator's suspicion.) [Note that this contradicts the conclusion from part (a).]

(c) Both tests require independent, random samples. The condition required for the t test but not for the Wilcoxon-Mann-Whitney test is that the population distributions are normal. The frequency distribution for the mechanical group is highly skewed, with some observations (2996 and 3452) that are much greater than the others. This casts doubt on the normality condition.

(d) K1 = 10 + 7 + 0 + 10 + 10 + 7 + 1 + 10 + 4 + 10 = 69 K2 = 3 + 2 + 1 + 2 + 2 + 3 + 5 + 5 + 5 + 3 = 31

where 1 denotes mechanical and 2 denotes manual.

7.90 (a) Let 1 denote control and let 2 denote stress.

H0: Stress has no effect on growth (µ1 = µ2) HA: Stress tends to retard growth (µ1 > µ2)

SE(y-1 - y-2) =2.132

13 + 1.732

13 = .7611.

ts = (30.59 - 27.78)/.7611 = 3.69. With df = 24, Table 4 gives t.005 = 2.797 and t.0005 = 3.745. (Formula (7.1) gives df = 23, which makes no appreciable change.) Thus, .0005 < P < .005, so we reject H0.

(b) There is sufficient evidence (.0005 < P < .005) to conclude that stress tends to retard plant growth.

7.91 Let 1 denote control and let 2 denote stress.

SE(y-1 - y-2) =2.132

13 + 1.732

13 = .7611.

(30.59 - 27.78) ± (2.064)(.7611) (df = 24)

(1.24,4.38) or 1.24 < µ1 - µ2 < 4.38 cm.

(a) The confidence interval indicates that the difference is at least 1 cm and so is "horticulturally important."

(b) According to the confidence interval, the difference could be greater or smaller than 2 cm, so the data do not indicate whether the difference is "horticulturally important."

115

(c) The confidence interval indicates that the difference is less than 5 cm and so is not "horticulturally important."


H0: Stress has no effect on growth HA: Stress tends to retard growth

The data are already arrayed in increasing order. We let Y1 denote control and Y2 denote stress. For the K1 count, we note that there is one Y2 less than the first Y1; there are ten Y2's less than the second Y1; there are twelve Y2's less than the third, fourth, fifth, sixth, and seventh Y1; there are twelve Y2's less than the eighth Y1 and one equal to it; and there are thirteen Y2's less than the ninth through thirteenth Y1. Thus,

K1 = 1 + 10 + 12 + 12 + 12 + 12 + 12 + 12.5 + 13 + 13 + 13 + 13 + 13 = 148.5.

For the K2 count, we note that there are no Y1's less than the first Y2; there is one Y1 less than the second through tenth Y2; there are two Y1's less than the eleventh and twelfth Y2; and there are seven Y1's less than the thirteenth Y2 and one equal to it. Thus,

K2 = 0 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2 + 2 + 7.5 = 20.5.

To check the counts, we verify that

K1 + K2 = 148.5 + 20.5 = 169 = (13)(13) = (n1)(n2).

To check the directionality of the data, we note that K1 > K2, which suggests a tendency for the Y1's to be larger than the Y2's, which would indicate that stress retards growth. Thus, the data do deviate from H0 is the direction specified by HA.

The Wilcoxon-Mann-Whitney test statistic is the larger of the two counts K1 and K2; thus Us = 148.5.

Looking in Table 6 under n = 13 and n' = 13, we find that for a directional alternative, the largest entry is 146 under the .0005 heading. Thus, the P-value is bracketed as

P < .0005. At significance level α = .01, we reject H0, since P < α. There is sufficient evidence (P < .0005) to

conclude that stress tends to retard growth.

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, Us = 80; the data deviate from H0 is 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 between .05 and .10, so we reject H0. There is sufficient evidence (.05 < P < .10) to conclude that biodiversity is greater along the Vermilion River than along the Black River.

7.94 Let 1 denote positive response and let 2 denote no response.

H0: Ovarian pH is not related to progesterone response HA: Ovarian pH is related to progesterone response

K1 = 1.5, K2 = 106.5, Us = 106.5 With n = 18, n' = 6, and a nondirectional alternative, the largest entry is 100 under the .0005 heading. Thus, the P-value is bracketed as

116

P < .0005. At significance level α = .05, we reject H0, since P < α. There is sufficient evidence (P < .0005) to

conclude that ovarian pH is lower among responders to progesterone than among nonresponders.

7.95 H0: Ovarian pH is not related to progesterone response (µ1 = µ2) HA: Ovarian pH is related to progesterone response (µ1 ≠ µ2)

SE(y-1 - y-2) =.1292

18 + .0812

6 = .04492.

ts = (7.373 - 7.708)/.04492 = -7.46. With df = 22, Table 4 gives t.0005 = 3.792. (Formula (7.1) gives df = 14.1, which makes no change in the conclusion.) Thus, P < .001, so we reject H0. There is sufficient evidence (P < .001) to conclude that ovarian pH is lower among responders to progesterone than among nonresponders.

7.96 The lack of a statistically significant difference in weight gain does not show that the new diet is as good

as the standard diet. (Evidence to this effect could be obtained from either a confidence interval or an analysis of the power of the test; see the next exercise.)

7.97 If the standard deviation is about 20% of the mean and the difference between the means is about 10%,

then

|µ1 = µ2|

σ = .10.20 = .5.

To achieve power of .50, Table 5 indicates that 32 animals per group would be required. Since 25 < 32, the power of this study was less than .50. Thus, the study did not have a reasonably good chance of detecting a population mean deficiency of 10% on the new diet.

7.98 (a) Two of the patients contributed two observations each to the data set. Thus, there is hierarchical

structure, so the t test is not appropriate.

(b) No. The Wilcoxon-Mann-Whitney test, like the t test, requires that the observations within a sample be independent of each other, so the Wilcoxon-Mann-Whitney test is not appropriate.

7.99 Let 1 denote low chromium and let 2 denote normal.

H0: Low chromium diet does not affect GITH (µ1 = µ2) HA: Low chromium diet does affect GITH (µ1 ≠ µ2)

y- 1 = 51.75, s1 = 5.526 y- 2 = 53.17, s1 = 4.123

SE(y-1 - y-2) =5.5262

14 + 4.1232

10 = 1.970.

ts = (51.75 - 53.17)/1.970 = -.72. Formula (7.1) gives df = 21.9; t(22).20 = .858, so P > .40. Using a computer, we get P = .48. Thus, we do not reject H0. There is insufficient evidence (P = .48) to conclude that low chromium diet affects GITH in rats.

7.100 Let 1 denote low chromium and let 2 denote normal.

H0: Low chromium diet does not affect GITH HA: Low chromium diet does affect GITH

117

Us = 70.5. n = 14, n' = 10, and a nondirectional alternative, the smallest entry is 93, under the .20 heading. Thus, P > .20 and we do not reject H0. There is insufficient evidence (P > .20) to conclude that low chromium diet affects GITH in rats.

7.101 (See Exercise 7.99 for basic computations.)

(a) (51.75 - 53.17) ± (2.074)(1.970) (df = 21.9 ≈ 22)

(-5.5,2.7) or -5.5 < µ1 - µ2 < 2.7 thousand cpm/gm.

(b) All values in the confidence interval are smaller in magnitude than 8 thousand cpm/gm; thus the data support the conclusion that the difference is "unimportant."

(c) The confidence interval indicates that the difference could be larger in magnitude than 4 thousand cpm/gm or smaller; thus the data do not indicate whether the difference is "unimportant."

7.102 (a) Let 1 denote infected and let 2 denote uninfected.

H0: Malaria is not related to stamina (µ1 = µ2) HA: Malaria is associated with decreased stamina (µ1 < µ2)

y- 1 = 26.87, s1 = 6.815; y- 2 = 32.23, s1 = 8.067.

SE(y-1 - y-2) =6.8152

15 + 8.0672

15 = 2.727.

ts = (26.87 - 32.23)/2.727 = -1.97. Formula (7.1) gives df = 27.2; t(27).03 = 1.963 and t(27).025 = 2.052, so .025 < P < .03. Using a computer, we get P = .0298. Thus, we reject H0. There is sufficient evidence (P = .0298) to conclude that malaria is associated with decreased stamina.

(b) H0: Malaria is not related to stamina HA: Malaria is associated with decreased stamina

Us = 155; the data deviate from H0 in the direction specified by HA. With n = 15, n' = 15, and a directional alternative, the entry under the .05 heading is 153 and the entry under the .025 heading is 161. Thus, .025 < P < .05 and we reject H0. There is sufficient evidence (.025 < P < .05) to conclude that malaria is associated with decreased stamina.

7.103 (a) Let 1 denote amphetamine and let 2 denote control.

H0: Amphetamine is not related to water consumption (µ1 = µ2) HA: Amphetamine is associated with decreased water consumption (µ1 < µ2)

y- 1 = 129.375, s1 = 27.850 y- 2 = 156.0, s1 = 25.322

SE(y-1 - y-2) =27.8502

4 + 25.3222

4 = 18.82.

ts = (129.375 - 156)/18.82 = -1.415. With df = n1 + n2 - 2 = 6 (Formula (7.1) yields df = 5.9), Table 4 gives t.20 = 0.906 and t.10 = 1.440, so .10 < P < .20. (Using a computer, we get P = .104.) Thus, we do not reject H0. There is insufficient evidence (10 < P < .20) to conclude that amphetamine is associated with decreased water consumption.

(b) H0: Amphetamine is not related to water consumption

118

HA: Amphetamine is associated with decreased water consumption

K1 = 4, K2 = 12, Us = 12; the data deviate from H0 in the direction specified by HA. With n = 4, n' = 4, and a directional alternative, the smallest entry is 13, under the .10 heading. Thus, P > .10 and we do not reject H0. There is insufficient evidence (P > .10) to conclude that amphetamine is associated with decreased water consumption.

7.104 (a) False. The confidence interval includes zero, so we are not confident that µ1 and µ2 are different.

(b) True. This is what a confidence interval tells us. (c) False. We know that y- 1 - y- 2 is exactly 6.9.

(d) False. The confidence interval is used to make an inference about the difference between µ1 and µ2; it does not tell us about individual data points (such as the length of hospitalization for a nitric oxide infant).

•7.105 False. The 95% confidence interval includes zero, which means that the P-value for a nondirectional

test is greater than .05. Thus, we would not reject H0 at the .05 significance level.

7.106 (a) False. The P-value for a test is the probability of getting data at least as extreme as those obtained, if H0 is true; it is not the probability that the null hypothesis is true.

(b) True. The P-value for a test is the probability of getting data at least as extreme as those obtained, if H0 is true, which is what this statement says.

(c) False. The probability that H0 is rejected depends on the power of the test, which is not known. (If H0 is true -- and we don't know if it is true or not -- and a new study is done that uses α = .04, then there is a 4% probability that H0 will be rejected.)

7.107 (a) False. The confidence interval gives a range that we infer covers µ1 - µ2. It does not tells us where

the bulk of the data lie.

(b) True. This is what a confidence interval tells us.

(c) False. The confidence interval is used to make an inference about the difference between µ1 and µ2; it does not tell us about future values of y- 1 and y- 2. (If µ1 - µ2 were exactly equal to the y- 1 - y- 2 difference obtained in this study -- which is not at all likely -- then the confidence interval could be used to predict where future y- 1 - y- 2 differences would fall.)

(d) False. The confidence interval is used to make an inference about the difference between µ1 and µ2; it does not tell us about individual data points (such as the number of calories in a particular entree).

119

7.108 (a) The normal probability plots show very mild skewness:

(b) The following computer output shows that the P-value is .0198.

(c) After taking logarithms of each observation, the normal probability plots look more linear:

The P-value for the t-test is now .0156, which is almost the same as the P-value before the transformation:

250

375

500

625

-1.25 0.00 1.25

nscores

250

375

500

625

-1.25 0.00 1.25

nscores

Far

Nea

r

2-Sample t-Test of µ1-µ2

Ho: µ1-µ2 = 0 Ha: µ1-µ2 ° 0near - far :

Test Ho: µ(near)-µ(far) = 0 vs Ha: µ(near)-µ(far) ° 0Difference Between Means = 69.029412 t-Statistic = 2.388 w/65 dfReject Ho at Alpha = 0.05p = 0.0198

2.500

2.625

2.750

-1.25 0.00 1.25

nscores

Log(

Nea

r)

2.375

2.500

2.625

2.750

-1.25 0.00 1.25

nscores

Log(

Far)

120

2-Sample t-Test of µ1-µ2

Ho: µ1-µ2 = 0 Ha: µ1-µ2 ° 0

Log(near) - Log(far) :

Test Ho: µ(Log(near))-µ(Log(far)) = 0 vs Ha: µ(Log(near))-µ(Log(far)) ° 0Difference Between Means = 0.07797664 t-Statistic = 2.483 w/65 dfReject Ho at Alpha = 0.05p = 0.0156

(d) When the sample sizes are fairly large the t-test gives almost exactly the same P-value for data that are

midly skewed as for data that are not skewed. That is, the t-test is robust.

7.109 (a) Let 1 denote andro and let 2 denote control.

H0: Andro has no effect (µ1 = µ2) HA: Andfo has an effect (µ1 ≠ µ2)

SE(y-1 - y-2) =13.32

9+

12.52

10= 5.94.

ts = (14.4 – 20.0)/5.94 = -.94. With df =16, Table 4 gives t.20 = 0.865 and t.10 = 1.337, so .20 < P < .40. (Using a computer, we get P = .359.) Thus, we do not reject H0. There is insufficient evidence (20 µ2)

The data deviate from H0 in the opposite direction from what HA predicts. Thus, P > .50 and we do not reject H0. There is insufficient evidence (P > .50) to conclude that andro increases lat pulldown strength.

7.110 A treatment was given to 244 subjects and the number of drinks for each subject was recorded for 7 days.

The treatment subjects were compared to 238 control subjects. The null hypothesis that the treatment has no effect on drink consumption was rejected in a one-sided test. There is strong evidence (P = .0031) that the treatment is associated with a reduction in drink consumption. Moreover, we can be 95% confident that on average treatment subjects consume between .92 and 5.56 fewer drinks, per 7 days, than do control subjects.

chapter 7 comparison of two independent samples · · 2006-04-10using df = 1000 (the nearest...

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