Trimester 3 Statistics Exam Review Answers
May 2008
  
*
1. Find n so that z * 
  1 .   8, z  1.9599 so n  245.86 . Sample size should
n


be 246.
2. a. A one sample t test is appropriate is the 10 fish are an SRS of the entire fish
population and if the PCB level in the population is normally distributed.
H0 :   5
, x  5.13, s  1.05, t  0.3907
Ha :   5
Df=9, P value =0.35.
c. If the mean PCB level in the population were 5 ppb, there is a 35% chance
that we would observe a sample with x  5.13 .
d. There is no evidence to indicate that the mean PCB concentration in the
lake is more than 5 ppb.
H 0 :   10
3. a. If  is the mean arsenic level in the local tap water, test
.
H a :   10
b. Type I: conclude that arsenic levels are over 10 ppm when in fact they are
not that high. This error would alarm residents unnecessarily. Type II:
conclude that arsenic levels are no high than 10 ppm when in face they are
higher. This error would give residents a false sense of security.
4. pm =proportion of males who recently skipped; p f =proportion of females who
recently skipped
a. 0.0153  p f  pm  0.145
b.
b. pˆ f  64 / 253  0.253 and pˆ m  71/ 411  0.173
Interval is pˆ f  pˆ m  z*
ME= z*
pˆ f 1  pˆ f 
nf
pˆ f 1  pˆ f 
nf


pˆ m 1  pˆ m 
and
nm
pˆ m 1  pˆ m 
0.253  0.747 0.173  0.827
 1.96

nm
253
411
ME=1.96*0.03309=0.06484
0.253  0.173
 2.494 ; P-value =0.013
c. z 
0.253  0.747 0.173  0.827

253
411
There is strong evidence of a difference between the proportions of males and females
who have recently skipped.
5 a. p1 =proportion of purchases that are Seabiscuit Oats; p2 =proportion of
purchases that are Secretariat Oats; p3 =proportion of purchases that are Whirlaway
H 0 : p1  0.5, p2  0.25, p3  0.25
Oats;
H a : at least one proportion not as claimed
b. Seabiscuit Oats = 100; Secretariat Oats = 50; Whirlaway Oats = 50
c. df=2
d. P value = -0.02367; there is strong evidence against the claim that
p1  0.5, p2  0.25, p3  0.25
956  450
 418.48
6 a. expected number who plead not guilty and are sent to prison =
1028
 418.48 31.52 
b. 
 ; df=1; chi-square = 42.556; P value =3.52x10^-11; reject null .
537.52 40.48
There is strong evidence against the claim that plea and sentence are independent.
7. a. y  24.56  31.69 x , 62.59 lbs.
b. 95% C.I. for slope: 20.83    42.55
H 0 :   35
31.69  35
 .70291 ; P value =0.25; there is not statistical
c.
; t value =
4.709
H a :  <35
evidence to reject the null hypothesis that the slope is 35
8. TRUE
9. TRUE
10. FALSE
11. FALSE
12. TRUE
13. FALSE
14. TRUE
15. FALSE
16. TRUE
Additional problems:
1.
1. Let  represent the true population proportion of uninjured occupants.
2. H o :   0.25
3. H a :   0.25
4.   0.01
5. Since n  = 319(0.25)  5, and n(1 -  ) = 319(0.75)  5, the large sample z test may
p  0.25
be used. z 
0.25(0.75)
n
95
6. n = 319, x = 95, p =
= 0.297806
319
.297806  0.25 0.047806
z

 1.97
0.024224
0.25(0.75)
319
7. P-value = area under the z curve to the right of 1.97 = 1-0.9756 – 0.0244
8. Since the P-value is less than  , H o is rejected. There is sufficient evidence in this
sample to support the conclusion that the true proportion of uninjured occupants exceeds
0.25.
2.
Let

represent the response rate when the distributor is stigmatized by an eye patch.
H o :   0.40
H a :   0.40
  0.05
The test statistic is: z 
p  0.40
0.40(0.60)
n
From the data: n  200, p 
109
 0.545
200
0.545  0.40
 4.19
0.40(0.60)
200
P-value = area under the z curve to the right of 4.19  1-1 = 0
z
Since the P-value is less than the  value of 0.05, H o is rejected. The data strongly
suggests that the response rate does exceed the rate in the past.
3.
1. Population characteristic of interest:  = true average fuel efficiency
2. H o :   30
3. H a :   30
4.   0.05 (for demonstration purposes)
X  30
5. Test statistic: t 
s
n
6. Assumptions: This test requires a random sample and either a large sample size
(generally n  30) or a normal population distribution. Since the sample size is only 6, we
can look at a box plost of the data. It shows symmetry, indicating that it would not be
unreasonable to assume that the population would be approximately normal. Hence we can
proceed with a t-test.
7. Computations: n = 6, X =29.33, s=1.41
29.33  30
t
 1.164
1.41
6
8. P-value=area under the 5 d.f. t curve to the left of -1.164 = 0.15
9. Conclusion: Since the P-value is greater than  , the null hypothesis is not rejected at
the 0.01 level. The data does not contradict the prior belief that the average fuel efficiency
is at least 30.
4.
Let 1 denote the mean number of goals scored per game for games in which Gretzky
played and  2 the mean number of goals scored per game for games in which he did not
play.
H 0 : 1  2  0
H a : 1  2  0
  0.01
Test statistic: t 
( X1  X 2 )  0
s12 s2 2

n1 n2
Assumptions: The population distributions are (at least approximately) normal and the two
samples are independently selected random samples.
n1  41, X 1  4.73, s1  1.29, n2  17, X 2  3.88, s2  1.18
t
(4.73  3.88)  0
2
(1.29) (1.18)

41
17
df 
2

 s12 s2 2 
 

 n1 n2 
0.85
 2.4286
0.3500
2
2
1  s12 
1  s2 


  
n1  1  n1  n2  1  n2 
2
2

(0.040588 0.081906) 2
 32.6
(0.040588) 2 (0.081906) 2

40
16
So df = 32 (rounded down to an interger)
P-value = area under the 32 df t curve to the right of 2.4286  0.0105
Since the P-value exceeds  , H o is not rejected. At a significance level of 0.01, the
sample data does not support the conclusion that the mean number of goals scored per
game is larger when Gretzky played than when he didn’t play.
5.
a. Let 1 denote the true average peak loudness for open-mouth chewing and  2 the true
average peak loudness for closed mouth chewing. Then 1  2 denotes the difference
between the means of open-mouthed and closed-mouth chewing.
n1  10, X 1  63, s1  13, n2  10, X 2  54, s2  16
V1 
s12 (13)2

 16.9
n1
10
V2 
s2 2 (13) 2

 25.6
n2
10
(V1  V2 ) 2
(16.9  25.6) 2
1806.25


 17.276
2
2
2
2
V1
V2
(16.9) (25.6)
104.552222


9
9
n1  1 n2  1
Use df = 17
df 
The 95% confidence interval for 1  2 based on this sample is
(63  54)  2.11 16.9  25.6  9  2.11(6.519202)  9  13.75  ( 4.75, 22.75)
Observe that the interval includes 0, and so 0 is one of the plausible values of 1  2 . That
is, it is plausible that there is no difference in the mean loudness for open-mouth and
closed-mouth chewing of potato chips.
b. Let 1 denote the true average peak loudness for closed-mouth chewing potato chips
and  2 the true average peak loudness for closed-mouth chewing tortilla chips.
H 0 : 1  2  0
H a : 1  2  0
  0.01
Test statistic: t 
( X1  X 2 )  0
s12 s2 2

n1 n2
Assumptions: the population distributions are (at least approximately) normal and the two
samples are independently selected random samples.
n1  10, X 1  54, s1  16, n2  10, X 2  53, s2  16
t
df 
(54  53)  0
(16) 2 (16)2

10
10

1.0
 0.1398
7.1554
 s12 s 2 2 
 

 n1 n2 
2
2
1  s12 
1  s2 


  
n1  1  n1  n2  1  n2 
2
2

(25.6  25.6) 2
 18
(25.6) 2 (25.6) 2

9
9
So df = 18 (rounded down to an integer)
P-value = 2(area under the 18 df t curve to the right of 0.1398)  2(0.445) = 0.890
Since the P-value exceeds  , H 0 is not rejected. There is not sufficient evidence to
conclude that there is a difference in the mean peak loudness for closed-mouth chewing of
tortilla chips and potato chips.
c. Let 1 denote the true average peak loudness for fresh tortilla chips when chewing
closed-mouth. Let  2 denote the true average peak loudness of stale tortilla chips when
chewing closed-mouth
H 0 : 1  2  0
H a : 1  2  0
  0.05
Test statistic: t 
( X1  X 2 )  0
s12 s2 2

n1 n2
Assumptions: The population distributions are (at least approximately) normal and the two
samples are independently selected random samples.
n1  10, X 1  56, s1  14, n2  10, X 2  53, s2  16
(56  53)  0
t
(14) 2 (16) 2

10
10
df 

3.0
 0.4462
6.723
 s12 s 2 2 
 

 n1 n2 
2
2
1  s12 
1  s2 


  
n1  1  n1  n2  1  n2 
2
2

(19.6  25.6) 2
 17.69
(19.6) 2 (25.6) 2

9
9
So df = 17 (rounded down to an interger)
P-value = area under the 17 df t curve to the right of 0.4462  0.331
Since the P-value exceeds  , H 0 is not rejected. There is not sufficient evidence to
conclude that there is a difference in the mean peak loudness when chewing fresh or stale
tortilla chips closed-mouth.
6.
H 0 : Seat location and motion sickness are independent.
H a : Seat location and motion sickness are not independent
  0.05
(observedcount  expectedcount )2
expectedcount
Observed and expected frequencies are given in the table below (expected frequencies in
parentheses)
Nausea
No nausea
Front
58
870
928
(118.85)
(809.15)
Seat Location
Middle
166
1163
1329
(170.21)
(1158.79)
Rear
193
806
999
(127.94)
(871.06)
417
2839
3256
Test statistic: X 2  
X2 
(58  118.85) 2 (166  170.21) 2 (193  127.94) 2 (870  809.15)2 (1163  1158.79) 2 (806  871.06) 2





 73.789
118.85
170.21
127.94
809.15
1158
871.06
Df=(3-1)(2-1) = 2. P-value < 0.001. Hence the null hypothesis is rejected. The data provide
strong evidence to conclude that seat location and nausea are dependent.