10.30 In Dallas, some fire trucks were painted yellow (instead of red) to heighten their
visibility. During a test period, the fleet of red fire trucks made 153,348 runs and had
20 accidents, while the fleet of yellow fire trucks made 135,035 runs and had 4
accidents. At a = .01, did the yellow fire trucks have a significantly lower accident
rate? (a) State the hypotheses. (b) State the decision rule and sketch it. (c) Find the
sample proportions and z test statistic. (d) Make a decision. (e) Find the p-value and
interpret it. (f) If statistically significant, do you think the difference is large enough to
be important? If so, to whom, and why? (g) Is the normality assumption fulfilled?
Explain.
Source: The Wall Street Journal, June 26, 1995, p. B1.
Accident Rate for Dallas Fire Trucks
Statistic Red Fire Trucks Yellow Fire Trucks
Number of accidents x1 = 20 accidents x2 = 4 accidents
Number of fire runs n1 = 153,348 runs n2 = 135,035 runs
Solution:
a) Null Hypothesis: H0: p1 = p2 vs. Ha: p1 > p2
b)
Rejection region can be defined as
z > z
Here significance level is 0.01,
So, z0.01 = 2.326 [critical value](Using Statistical Ratio Calculator
From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.01
significance)
Reject H0 if z > 2.326
c)
pˆ 
X1  X 2
20  4

 8.322 *10 5
n1  n 2
153,348  135,035
z
pˆ 1  pˆ 2
 1
1 

pˆ 1  pˆ  
 n1 n 2 
20
4

153,348 135,035


8.322 * 10 5 1  8.322 * 10 5
1
1





 153,348 135,035 

 2.961
d)
Since calculated value of test-statistics 2.961 is greater than 2.326, we can reject the null
hypothesis and conclude that yellow fire trucks have a significantly lower accident rate.
e) Calculating p-value:
Degree of freedom = DF = 153348+135035 -2 = 288381
P(z288381 > 2.961) = 0.001533 (Using http://www.danielsoper.com/statcalc/calc08.aspx
With t = 2.961 and DF = 288381)
Calculated value of p is 0.001533 which is less than significant level 0.01. Hence we can
reject the null hypothesis.
f) Yes, the difference is large enough to be important for the selected sample at significant level
0.01.
g) The normality assumption fulfilled won’t be fulfilled because second sample is very small
consisting only of 4 accidents.
10.44 Does lovastatin (a cholesterol-lowering drug) reduce the risk of heart attack? In a
Texas study, researchers gave lovastatin to 2,325 people and an inactive substitute to 2,081
people (average age 58). After 5 years, 57 of the lovastatin group had suffered a heart
attack, compared with 97 for the inactive pill. (a) State the appropriate hypotheses. (b)
Obtain a test statistic and p-value. Interpret the results at a = .01. (c) Is normality assured?
(d) Is the difference large enough to be important? (e) What else would medical researchers
need to know before prescribing this drug widely? (Data are from Science News 153 [May
30, 1998], p. 343.)
Solution:
a) Null Hypothesis: H0: p1 = p2 vs. Ha: p1 < p2
b)
pˆ 
z
X1  X 2
57  97

 0.035
n1 n 2
2,325  2,081
pˆ 1  pˆ 2
1
1 
pˆ 1  pˆ   
 n1 n 2 

57
97

2,325 2,081
1 
 1
0.0351  0.035


 2,325 2,081 
 3.984
Calculating p-value:
Degree of freedom = DF = 2325+2081 -2 = 4404
P(z4404 < -3.984) = 0.000034 (Using http://www.danielsoper.com/statcalc/calc08.aspx
With t = -3.984 and DF = 4404)
Calculated value of p is 0. 000034 which is less than significant level 0.01. Hence we can
reject the null hypothesis and conclude that the lovastatin reduces the risk.
c) Since sample sizes are large enough, we can assume that it is normality distribution.
d) Yes, difference is large enough to be important.
e) Researchers should verify the sampling strategy before reaching to conclusion that it has been
collected using random sampling. They should also test the side effects of drug before
prescribing it widely.
10.46 To test the hypothesis that students who finish an exam first get better grades,
Professor Hardtack kept track of the order in which papers were handed in. The first 25
papers showed a mean score of 77.1 with a standard deviation of 19.6, while the last 24
papers handed in showed a mean score of 69.3 with a standard deviation of 24.9. Is this a
significant difference at a = .05? (a) State the hypotheses for a right-tailed test. (b) Obtain a
test statistic and p-value assuming equal variances. Interpret these results. (c) Is the
difference in mean scores large enough to be important? (d) Is it reasonable to assume
equal variances? (e) Carry out a formal test for equal variances at a = .05, showing all steps
clearly.
Solution:
(a) Null Hypothesis: H0: (1 - 2) =0 vs. Ha: (1 - 2) 0
(b) Calculating test-statistics,
S
t
(n1  1) S 12  (n 2  1) S 22
(25  1)19.6 2  (24  1)24.9 2

 22.351
n1  n 2  2
25  24  2
x1  x 2
1
1
S

n1 n 2

77.1  69.3
 1.221
1
1
22.351

25 24
Calculating p-value:
Degree of freedom = DF = 25+24 -2 = 47
P(z47 < 1.221) = 0.114088 (Using http://www.danielsoper.com/statcalc/calc08.aspx
With t = 1.221 and DF = 47)
Calculated value of p is 0. 114088 which is greater than significant level 0.05. Hence we fail
to reject the null hypothesis and we don’t have sufficient evidence to conclude that students
who finish an exam first get better grades.
(c) The difference in mean scores is not large enough to be important since calculated value
of p is 0. 114088 which is greater than significant level 0.05.
(d) It is reasonable to assume equal variances as students are from same class which can have
similar standard deviation for the case of students who submit their papers and who submit their
papers late.
(e) Formal test for equal variance,
Null Hypothesis: Ho :  12   22  0 vs. Ha :  12   22  0
Calculated value of F,
s12 24.9 2
F 2 
 1.614
s 2 19.6 2
Critical value of F,
Probability = 0.025 DFn=24 DFd=23
F= 2.2989
2.50% of a F distribution has values greater than 2.299.
(Using http://www.graphpad.com/quickcalcs/StatRatio2.cfm )
Calculating p-value,
F=1.614 DFn=24 DFd=23
The P value equals 0.1276
By conventional criteria, this difference is considered to be not statistically significant.
Since calculated p-value is greater than significant level 0.05, we fail to reject the null hypothesis
which states that two populations have equal variances.
10.56 A sample of 25 concession stand purchases at the October 22 matinee of Bride of
Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the
October 26 evening showing of the same movie, for a sample of 25 purchases the mean was
$5.12 with a standard deviation of $2.14. The means appear to be very close, but not the
variances. At a = .05, is there a difference in variances? Show all steps clearly, including an
illustration of the decision rule.
Solution:
Null Hypothesis: Ho :  12   22  0 vs. Ha :  12   22  0
Calculated value of F,
F = (σ1)²/(σ2)² = 3.02²/2.14² = 1.99
Critical value of F,
Probability = 0.025 DFn=24 DFd=24
F= 2.2693
2.50% of a F distribution has values greater than 2.269.
(Using http://www.graphpad.com/quickcalcs/StatRatio2.cfm )
Calculating p-value,
F=1.99 DFn=24 DFd=24
The P value equals 0.0492
By conventional criteria, this difference is considered to be statistically significant.
Since calculated value of F is 1.99 which is less than critical value 2.2693, we fail to reject the
null hypothesis and can’t conclude that there is a difference in variances.
11.24 In a bumper test, three types of autos were deliberately crashed into a barrier at 5
mph, and the resulting damage (in dollars) was estimated. Five test vehicles of each type
were crashed, with the results shown below. Research question: Are the mean crash
damages the same for these three vehicles? Crash1
Crash Damages
Goliath
1,600
760
880
1,950
1,220
Varmint
1,290
1,400
1,390
1,850
950
Weasel
1,090
2,100
1,830
1,250
1,920
Solution:
Result of ANOVA.
Anova: Single
Factor
SUMMARY
Groups
Column 1
Column 2
Column 3
Count
5
5
5
Sum
Average Variance
6410
1282
246320
6880
1376
103580
8190
1638
195170
ANOVA
Source of Variation
Between Groups
Within Groups
SS
340360
2180280
Total
2520640
df
2
12
MS
170180
181690
F
P-value
F crit
0.93665 0.418802 3.885294
14
(i) Null Hypothesis: H0: 1 = 2 = 3 vs. Ha: (1 = 2 != 3) or (1 != 2 = 3) or (1 != 2 != 3)
(ii) Reject H0 if z > 2.326 if calculated p-value < 0.05
(iii) Since calculated value of p in ANOVA table is 0.4188 which is greater than significant level
0.05 , we fail to reject the null-hypothesis and can’t conclude that the mean crash damages are
different for these three vehicles.