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## Finance.cycu.edu.tw

**Chapter 22 Solutions **
**22.1. (a) **Diagram below.

**(b) **The null hypothesis is “all groups have the same mean rest period,”

and the alternative is “at least one group has a different mean rest period.” The

*P*-value shows significant evidence against

*H*0, and the graph leads us to conclude that caffeine has the effect of reducing the length of the rest period.

**Note: ***Students* *might* *be* *tempted* *to* *think* *that* *the* *alternative* *hypothesis* *is* *“mean *

rest *period* *decreases* *with* *increasing* *caffeine* *dosage”* *or* *something* *similar,* *but* *ANOVA *

alternatives *are* *nondirectional. *
**22.2. (a) **The null hypothesis is “all age groups have the same (population) mean road-rage

measurement,” and the alternative is “at least one group has a different mean.”

**(b) **The

*F* test is

quite significant, giving strong evidence that the means are different. The sample means suggest

that the degree of road rage decreases with age. (We assume that higher numbers indicate

*more* road rage.)

**22.3. (a) **The stemplots (right) appear to suggest that

logging reduces the number of trees per plot and
that recovery is slow (the 1-year-after and 8-years-
after stemplots are similar).

**(b) **The means lead one

to the same conclusion as in (a).

**(c) **In testing

*H*
=

*µ*2 =

*µ*3 vs.

*Ha*: not all means are the same, we find that

*F* =11.43 with df 2 and 30, which
has

*P* = 0.000205, so we conclude that these differences are significant: The number of trees per

**22.4. (a) **The stemplots (right) show no extreme

outliers or skewness.

**(b) **The means suggest that a

dog reduces heart rate, but being with a friend
appears to raise it.

**(c) ***F* = 14.08 and

*P* = 0.000

(meaning

*P* < 0.0005), which means we reject

*H*
*µP* =

*µF* =

*µC* in favor of

*Ha*: at least one mean is
different. Based on the confidence intervals, it
appears that the mean heart rate is lowest when a
pet is present (although this interval overlaps the
control interval) and is highest when a friend is
present (although again, this interval overlaps the control interval).

**22.5. (a) ***I*, the number of populations, is 3; the sample sizes from each population are

*n*1 =

*n*2 = 12

and

*n*3 = 9; the total sample size is

*N* = 33.

**(b) **Numerator (“Group”):

*I* – 1 = 2, denominator

(“Error”):

*N* –

*I* = 30.

**(c) **Because

*F* >9.22, the largest critical value for an

*F* 2,25 distribution in

Table D, we conclude that

*P* <0.001.

**22.6. (a) ***I*, the number of populations, is 3; the sample sizes from each population are

*n*1 =

*n*2 =

*n*3 =

15; the total sample size is

*N* = 45.

**(b) **Numerator (“Group”):

*I* – 1 = 2, denominator (“Error”):

*N* –

*I* = 42.

**(c) **Because

*F* >9.22, the largest critical value for an

*F* (2,25) distribution in Table D, we

conclude that

*P* < 0.001.

**22.7. (a) **We have

*I* = 4 populations; individual sample size

*n*1 = 146,

*n*2 = 125,

*n*3 = 104, and

*n*4 =

84; and total sample size

*N* = 459. The degrees of freedom are therefore 3 and 455.

**(b) **Because

3.18 <

*F* < 3.88, 0.010 <

*P* < 0.025.

**(c) **Because

*F* < 2.11,

*P* > 0.100.

**(d) **Because 5.63 <

*F*,

*P* <0.001.

**22.8. (a)** Yes: largest

*s*
**(b) **Yes: largest

*s*
**22.9. **The standard deviations (0.1201, 0.1472, 0.1134)

do not violate our rule of thumb. However, the
distributions appear to be skewed and have outliers,

*Chapter 22 One-Way Analysis of Variance: Comparing Several Means*
**22.10. (a) **The biggest difference is that single men earn considerably less than men who have been

or are married. Widowed and married men earn the most; divorced men earn about $1300 less (on

the average), and single men are $4000 below that.

**(b) **Yes: 8119 1.42.

**(c) **The degrees of

freedom are 3 and 8231.

**(d) **The sample sizes are so large that even

*small* differences would be

found to be significant; these differences are fairly large.

**(e) **No: Single men are likely to be

younger than men in the other categories. This means that typically they have less experience, and

have been with their companies less time than the others, and so have not received as many raises,

etc. (Age is the lurking variable.)

**22.11. (a) **The sample sizes are quite large, and the

*F* test is robust against non-Normality with large

samples.

**(b) **Yes (barely): The ratio is 3.11 = 1.94.

**(c) **We have

*I* = 3 and

*N* = 1342. The details of

the computations are given here; the Minitab output below confirms the computed values.
SSG = 244(2.22 −1.31) + 734(1.33 −1.31) + 364(0.66 −1.31)
SSE = 243× 3.11 + 733× 2.21 + 363×1.60

**(d) **With df 2 and 1339, we find that

*P* < 0.001; this is strong evidence that the means differ

**Minitab output **
**22.12. (a) **Yes: The rule-of-thumb ratio is 5.2 = 1.24.

**(b) ***x *= 9.92, and MSG = 10.0 (details

below).

**(c) **MSE = 21.9 (details below).

**(d) ***F* = 0.46 (details below). With df 2 and 112 (use 2

and 100 in the table), we find

*P* > 0.100, so we have no reason to doubt the null hypothesis; that

is, there is not enough evidence to conclude that mean weight loss differs between these exercise

programs.

In the details of the computations, we have

*I* = 3 and

*N* = 115:
37(10.2−9.92) +36(9.3−9.92) +42(10.2−9.92)

**22.13. (a) **Multiply by

*n *to find the standard deviations:

*s*cold = 8.08 16 = 32.32,

*s*neutral

= 5.61 38 = 34.58, and

*s*hot = 4.10 75 = 35.51. This easily satisfies our rule of thumb:
35.51 =1.10.

** (b) **We have

*I* = 3 and

*N* = 129, so

(16)(28.89)−32.045) +(38)(32.93−32.045) +(75)(32.27−32.045)
With df 2 and 126, we find

*P* > 0.100, so we have no reason to doubt the null hypothesis; that is, there is no evidence that nest temperature affects mean weight.

**22.14. (a) **Multiply by

*n * to find the standard deviations:

*s*cold = 5.67 16 = 22.68,

*s*neutral

= 4.24 38 = 26.14, and

*s*hot = 2.70 75 = 23.38. This easily satisfies our rule of thumb:
26.14 =1.15.

**(b) **We have

*I* = 3 and

*N* = 129, so

(16)(6.40−5.008) +(38)(5.82−5.008) +(75)(4.30−5.008)
With df 2 and 126, we find

*P* > 0.100, so we have no reason to doubt the null hypothesis; that is, there is no evidence that nest temperature affects propensity to strike.

**22.15. **Populations: nonsmokers, moderate smokers, and heavy smokers; response variable: hours of

sleep per night.

*I* = 3,

*n*1 =

*n*2 =

*n*3 = 200, and

*N* = 600, so there are

*I* – 1 = 2 and

*N* –

*I* = 597 degrees of freedom.

**22.16. **Populations: consumers (responding to different package designs); response variable:

attractiveness rating.

*I* = 6,

*n*1 =

*n*2 =

*n*3 =

*n*4 =

*n*5 =

*n*6 = 120, and

*N* = 720, so there are

*I* – 1 = 5 and

*N* –

*I* = 714 degrees of freedom.

**22.17. **Populations: tomato varieties; response variable: yield.

*I* = 4,

*n*1 =

*n*2 =

*n*3 =

*n*4 = 10, and

*N* =

40, so there are

*I* – 1 = 3 and

*N* –

*I* = 36 degrees of freedom.

**22.18. **Populations: different concrete mixtures; response variable: strength.

*I* = 5,

*n*1 = · · · =

*n*5 = 6,

and

*N* = 30, so there are

*I* – 1 = 4 and

*N* –

*I* = 25 degrees of freedom.

**22.19. **Populations: students taught by different methods; response variable: test scores.

*I* = 4,

*n*1 =

*n*2 =

*n*3 = 10,

*n*4 = 12, and

*N* = 42, so there are

*I* – 1 = 3 and

*N* –

*I* = 38 degrees of freedom.

*Chapter 22 One-Way Analysis of Variance: Comparing Several Means*
**22.20. (a) **We test

*H*0:

*µ*1 =

*µ*2 =

*µ*3 =

*µ*4 =

*µ*5 vs.

*Ha*: not all means are the same.

**(b) ***N* = 168,

*I* = 5,

*n*1 = 39,

*n*2 = 35,

*n*3 = 29,

*n*4 = 30, and

*n*5 = 35.

**(c) **We have 4 and 163 degrees of freedom.

**22.21. (a) **The graph does suggest that emissions rise

when a plant is attacked, because the mean control

emission rate is half the smallest of the other rates.

**(b) **The null hypothesis is “all groups have the same

mean emission rate.” The alternative is “at least one

group has a different mean emission rate.”

**(c) **The

most important piece of additional information

would be whether the data are sufficiently close to

Normally distributed. (From the description, it seems

reasonably safe to assume that these are more-or-less

random samples.)

**(d) **The SEM equals

*s */ 8, so we can find the standard deviations by

multiplying by 8 — however, this factor of 8 would cancel out in the process of finding the ratio of the largest and smallest standard deviations, so we can simply find this ratio directly from the SEMs: 8.75

**22.22. (a) **The means are given in the table (below) and are shown in the plot (as the symbol “+”).

(The plot also shows the original data as solid circles.) The means vary over time but do not

consistently decrease.

**(b) **The standard deviations are also given in the table; the largest-to-

smallest ratio is 16.0873 3.49. Because this is much more than 2, ANOVA should not be used.

**(c) **The two-sample

*t* test does not require that the standard deviations be equal, but the ANOVA

assumes that they are and is not reliable if there is evidence that they are different. These data

suggest that variability increases over time.

**22.23. **Only Design A would allow use of one-way ANOVA because it produces four independent

sets of numbers. The data resulting from Design B would be dependent (a subject’s responses to the first list would be related to that same subject’s responses to the other lists), so that ANOVA would not be appropriate for comparison.

**22.24. **The ANOVA test statistic is

*F* = 4.92 (df 3 and 92), which has

*P* = 0.003, so there is strong

evidence that the means are not all the same. In particular, list 1 seems to be the easiest, and lists 3 and 4 are the most difficult.

**22.25. (a) **Stemplots (below) suggest that yields first increase with plant density, then decrease. The

standard deviation ratio is 1.95. With such small samples, outliers and skewness cannot be

assessed.

**(b) **We test

*H*0:

*µ*1 =

*µ*2 =

*µ*3 =

*µ*4 =

*µ*5 (all plant densities give the same mean yield per

acre) vs.

*H*a: not all means are the same. Minitab output is below; with

*F* = 0.50 and

*P* = 0.736,

we conclude that the differences are not significant.

**(c) **The sample sizes were small, which

means there is a lot of potential variation in the outcome. (That is why the confidence intervals

shown are very wide.)

**Minitab output **
Individual 95% CIs For Mean Based on Pooled StDev
4 131.03 18.09 (-----------*-----------)
3 143.07 11.44 (------------*-------------)
22.27 (---------------*----------------)

**22.26. (a) **The table is given in the Minitab output below; because 4.500 = 1.28, ANOVA should be

safe. The means appear to suggest that logging reduces the number of species per plot and that

recovery takes more than 8 years.

**(a) **ANOVA gives

*F* = 6.02 with df 2 and 30, so

*P* < 0.010

(software gives 0.006), so we conclude that these differences are significant; the number of

species per plot really is lower in logged areas.

**Minitab output **
Individual 95% CIs For Mean Based on Pooled StDev

*Chapter 22 One-Way Analysis of Variance: Comparing Several Means*
**22.27.(a) **Table below (part of Minitab out-put); stemplots at

right. The data suggest that the presence of too many
nematodes reduces growth.

**(b) **H0:

*µ*1 = · · · =

*µ*4 (all mean

heights are the same) vs.

*Ha*: not all means are the same.
This ANOVA tests whether nematodes affect mean plant
growth.

**(c) **Minitab output is shown below:

*F* = 12.08 df 3

and 12,

*P* = 0.001, so the differences are significant. The
first two levels (0 and 1000 nematodes) do not appear to be
significantly different, nor do the last two. However, it does 11
appear that somewhere between 1000 and 5000 nematodes
the effects of the worms and are hurt by their presence.

**Minitab output **

Analysis of Variance on Growth

Individual 95% CIs For Mean Based on Pooled StDev

**22.28. **We have

*I* = 3 and

*N* = 259. The details of the computations are given here; the Minitab

output that follows confirms the computed values.
SSG = 57(26,470 − 29,484) +108(30, 610 − 29,484)
SSE = 56 × 9507 +107 × 4504 + 93× 7158
With df 2 and 256 (use 2 and 200 in the table), we find

*P* < 0.001, so we have fairly strong

**Minitab output **
**22.29. **We have

*I* = 4 and

*N* = 32. Also, all SEMs must be squared and multiplied by 8 to find the

variances. The details of the computations are given here; the Minitab output below confirms the computed values.
SSG = 8(9.22 − 21.585) + 8(31.03 − 21.585)
With df 3 and 28, we find

*P* > 0.100, so we have no reason to doubt the null hypothesis; that is,
there is not enough evidence to conclude that mean emission rates are different.

**Minitab output **
*Chapter 22 One-Way Analysis of Variance: Comparing Several Means*
**22.30. (a) **The standard error is SE

= 0.38 /13 + 0.58 /17 = 0.1758, so the

*t* statistic is
− 34, which is obviously not significant. The conservative method gives df = 12,
and software reports

*P* = 0.74.

**(b) **The details of the computation are given here; Minitab output

is shown below.

With df 1 and 28, software reports that

*P* = 0.749.

**(c) **The two P-values are almost identical.

**Minitab output **
– – – – – – – – – – – – – – –

**T test **– – – – – – – – – – – – – – –

95% C.I. for mu Stock – mu Mutual: ( -0.42, 0.30) T–Test mu Stock = mu Mutual (vs not =): T= –0.34 P=0.74 DF= 27
– – – – – – – – – – – – – – –

**ANOVA **– – – – – – – – – – – – – – –

Analysis of Variance

**22.31. (a) **A chi-square test is appropriate, because we are comparing three proportions (attrition

rates in each of three groups).

**(b) **ANOVA is appropriate, because we are comparing three means

(weight loss in each of three groups).

**(c) **ANOVA is appropriate, because we are comparing three

means (duration of exercise in each of three groups).

**22.32. (a) **By moving the middle mean to the same level as the other two, it is possible to reduce

*F*
to 0.0236, which has a

*P*-value very close to the left end of the scale (near 1).

**(b) **By moving any

mean about 1 centimeter up or down (or any two means about 0.5 cm in opposite directions), the

value of

*F* increases (and

*P* decreases) until it appears at the right end of the scale.

**22.33. (a) ***F* can be made as small as 0.3174, while

*P* > 0.5.

**(b) ***F* can be made quite large (and

*P*
Source: http://finance.cycu.edu.tw/www/96statistics/SOLUTION-22.PDF

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