Elementary Probability & Statistics
Practice Exam #3
(10 points) 1. Thirty five randomly selected students took the statistics final. If the sample
mean was 80 and the population standard deviation was 12, construct the 97% confidence
interval for the mean score of all students. Assume that the population has a normal
distribution. Be sure to interpret your result and show your graph.
n(number of sample)
x
s or 
(Circle one)
Degree of Confidence
Critical Value (z/2or t/2)
35
80
12
97%
2.17
(Circle one)
Margin of Error (E)
E  2.17 
12
 4.4
35
Confidence Interval
Interpretation of confidence
interval.
75.6<  <84.4
We are 97% confident that interval from
75.6 to 84.4 actually does contain true
value 
Graph
1
2. The amounts (in ounces) of juice in eight randomly selected juice bottles are
14.1
15.1
14.6
15.2
14.7
15.5
14.8
15.3
Find a 90% confidence interval for the population standard deviation σ. Assume that the
population has a normal distribution. Be sure to interpret your result and show your
graph.
n(number of sample)
8
x
14.91
s or 
(Circle one)
0.45
Degree of Confidence
90%
Degree of Freedom (if required)
7
Critical Values (2L and 2R)
 L2  2.167,  R2  14.067
Confidence Interval
0.317    0.809
Interpretation of confidence
interval.
We are 90% confident that the interval from
0.317 to 0.809 does actually contain true
value 
Graph
2
3. Scores on a certain test are normally distributed with a variance of 82. A researcher
wishes to estimate the mean score achieved by all adults on the test. Find the sample size
needed to assure with 98% confidence that the sample mean will not differ from the
population mean by more than two units.
2
 (2.33)(9.06) 
n
  112
2

4. A random sample of 100 babies is obtained, and the mean head circumference is found
to be 40.6 cm. Assuming that the population standard deviation is known to be 1.6 cm,
use a 0.05 significance level to test the claim that the mean head circumference of all
two-month-old babies is equal to 40 cm. Use P  value method.
Original claim:   40
Competing idea:   40
Null hypothesis ( H 0 ):   40
Alternative hypothesis ( H 1 ):   40
Which tail test? Two tail test
Significance Level ( ) :0.05
Critical value(s) (if required): z  1.96,1.96
Test Statistic: z t .s. 
40.6  40
 3.75
1. 6
100
P  value : 2( p( z  3.75)  21  p( z  3.75)]  0.0002
Graph:
Final Conclusion in terms of null hypothesis ( H 0 ): Reject H 0
Final Conclusion in terms of original claim: There is sufficient evidence to warrant
rejection of the claim that   40 .
3
5. The Regional Transit claims that the average wait time is no more than 17 minutes in
any particular route. You survey 100 randomly selected passengers and find that they
waited an average of 18.4 minutes, with standard deviation 4.6 minutes. At the 0.01 level
of significance, what if anything could you conclude? Use traditional method.
Original claim:   17
Competing idea:   17
Null hypothesis ( H 0 ):   17
Alternative hypothesis ( H 1 ):   17
Significance Level ( ) : 0.01
Critical value(s): t  2.364
Right tail test
Test Statistic: t t .s. 
18.4  17
 3.06
4 .6
100
Graph:
Final Conclusion in terms of null hypothesis: Reject null hypothesis H 0
Final Conclusion in terms of original claim: There is sufficient evidence to warrant
rejection of the claim that the average wait time is no more than 17 minutes in a
particular route.
4