Problem Set
Problem 9.38 (p. 385)
A volunteer working at an animal shelter conducted a study on the effect of catnip on cats at the
shelter.
She recorded the number of “negative interactions” each of the cats made in 15 minute
periods before and after being given a teaspoon of catnip.
The paired measurements were collected
on the same day within 30 minutes of one another;
the data are given in the accompanying table:
Cat
Before (
1
Y
)
After (
2
Y
)
Difference
Amelia
0
0
0
Bathsheba
3
6
3
Boris
3
4
1
Frank
0
1
1
Jupiter
0
0
0
Lupine
4
5
1
Madonna
1
3
2
Michelangelo
2
1
1
Oregano
3
5
2
Phantom
5
7
2
Posh
1
0
1
Sawyer
0
1
1
Scary
3
5
2
Slater
0
2
2
Tucker
2
2
0
Mean
1.8
2.8
1
SD
1.66
2.37
1.20
(a)
Construct a 95% confidence interval for the difference in mean number of negative interactions.
1
2
1,
1.2,
0.3098
14
1
(2.145)(0.3098) :
1.66
0.34
d
d
s
SE
df
μ
μ
= 
=
=
=


<

< 
(b)
Construct a 95% confidence interval the wrong way, using the independent sample method.
How
does this interval differ from the one in part (a)?
2
2
1
2
1.66 /15
2.37 /15
0.747
14(
use the complicated formula, or df=28)
1
(2.145)(0.747) :
2.603
0.603
SE
df
or
μ
μ
=
+
=
=


<

<
The interval much wider than in (a).
Problem 9.39 (p. 385)
Refer to exercise 9.38.
Compare the before and after populations using a ttest at
0.05
α
=
.
Use a
nondirectional alternative.
µ1 mean number of negative interactions in cats
µ2 mean number of negative interactions in cats after catnip
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H0: The means of negative interactions before and after catnip are the same( µ1=µ2)
HA: They are different. ( µ1≠µ2)
Use paired nondirectional ttest with df=14.
Reject H0 if ts>2.145.
SE=1.2/√15=0.3098, ts=3.23, so reject H0.
Catnip changes (increases) the mean number of negative interactions in cats at significance level 0.05. (P<0.01)
Problem 9.40 (p. 385)
Refer to Exercise 9.38.
(a)
Compare the before and after populations using a sign test at
0.05
α
=
.
Use a nondirectional
alternative.
(b)
Calculate the exact Pvalue for the analysis in part (a)
Let p be the probability that the count of negative interactions is higher before catnip.
H0: p=0.5 Probability of having more negative interactions is the same before and after catnip.
HA: p≠0.5 The probabilities are different.
Use nondirectional sign test with n=12. Has folded binomial distribution with n=12, p=0.5 under
H0.
Will reject H0 if Bs<=critical value of Table 7=10.
N+=2, N=10, Bs=10<=10 so reject H0 at 0.05 significance level.
At 0.05 significance level the counts of negative interactions are different (higher) after catnip.
P=2x(0.5)^12[66+12+1]=0.0385 (this is the exact Pvalue in (a))
Problem 10.4 (p. 400)
At a Midwestern hospital there were a total of 932 births in 20 consecutive weeks.
Of these births, 216
occurred on weekends.
Do these data reveal more than chance deviation from random timing of the
births?
(Test for goodness of fit, with two categories of birth – weekday and weekend.
Use a
nondirectional alternative and let
α
= 0.05.)
If the timing of births is random, then 2/7 of all births should occur on weekends, and 5/7 on weekdays.
Do
births deviate from this random timing?
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 Fall '08
 BUD
 Statistics, Linear Regression, Normal Distribution, Regression Analysis, Statistical hypothesis testing, laetisaric acid

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