Data from the Motor Vehicle Department indicate that 80% of all licensed drivers are
older than age 25. a. In a sample of n=60 people who recently received speeding
tickets, 38 were older than 25 years and the other 22 were age 25 or younger. Is the
age distribution for this sample significantly different from the distribution for the
population of licensed drivers? Use a = .05.
Assumptions:
The 2 categories we are comparing are: Licensed drivers age 25 or younger, and licensed drivers
over age 25. Multiplying each of the expected proportions from the null hypothesis below by the
sample size of 60, we find that the expected counts are 12, and 48 respectively. Because each
expected count is at least 5, we can use the Goodness of Fit test.
Using the P-value Method
1. The claim is that the proportion of ages from this sample differ from p25  0.20 ,
p25  0.80 as reported by the DMV. The correct hypothesis is,
H 0 : p25  0.20 and p25  0.80
H1 : At least one proportion differs from above (claim)
This is a right-tailed test.
2. There are 2 categories that we are comparing. We must compute the Observed and
Expected count for each of the 2 categories. The Observed counts are the actual number
of counts from the sample. The Expected counts are found by multiplying the expected
proportion by the sample size. It will be helpful to make a table.
Age of
Driver
25 or Younger
Over 25
Observed
(O)
22
38
Expected
(E)
0.20*60=12
0.80*60=48
O  E 
E
2
 22  12 
12
2
 8.3333
 38  48
48
2
 2.0833
The test value is,
(O  E )2
2  
 8.333  2.0833  10.4166
E
Total
 10.4166
3. The P-value for a right-tailed  2 test with degrees of freedom = the number of categories minus
1 = 2 – 1 = 1 is found by finding the area to the right of the test value using a  2 distribution.
The P-value is  0.001249 using a calculator.
4. We are given a level of significance of   0.05 . Because the P-value is much smaller than
  0.05 , we Reject H 0 .
5. Because we Reject H 0 , we are finding evidence in support of the claim. Therefore, based on our
sample, it appears as though the proportion of ages of licensed drivers differ from p 25  0.20 ,
p25  0.80 as reported by the DMV.
Same as above, the total number of participants is 60. Obtain the expected frequencies using 80% and
20%. b. In a sample of n=60 people who recently received parking tickets, 43 were older than 25 years
and the other 17 were age 25 or younger. Is the age distribution for this sample significantly different
from the distribution for the population of licensed drivers? Use a = .05.Same as above only the
observed frequencies changed
Assumptions:
The 2 categories we are comparing are: People receiving parking tickets age 25 or younger, and
people receiving parking tickets over age 25. Multiplying each of the expected proportions from
the null hypothesis below by the sample size of 60, we find that the expected counts are 12, and
48 respectively. Because each expected count is at least 5, we can use the Goodness of Fit test.
Using the P-value Method
1. The claim is that the proportion of ages from this sample differ from p25  0.20 ,
p25  0.80 as reported by the DMV. The correct hypothesis is,
H 0 : p25  0.20 and p25  0.80
H1 : At least one proportion differs from above (claim)
This is a right-tailed test.
2. There are 2 categories that we are comparing. We must compute the Observed and
Expected count for each of the 2 categories. The Observed counts are the actual number
of counts from the sample. The Expected counts are found by multiplying the expected
proportion by the sample size. It will be helpful to make a table.
Age of
Person
Receiving
Parking
Ticket
25 or Younger
Over 25
Observed
(O)
17
43
Expected
(E)
0.20*60=12
0.80*60=48
O  E 
E
2
17  12 
12
2
 2.0833
 43  48
48
2
 0.5208
Total
 2.6041
The test value is,
(O  E ) 2
2  
 2.0833  0.5208  2.6041
E
3. The P-value for a right-tailed  2 test with degrees of freedom = the number of categories minus
1 = 2 – 1 = 1 is found by finding the area to the right of the test value using a  2 distribution.
The P-value is  0.1066 using a calculator.
4. We are given a level of significance of   0.05 . Because the P-value is larger than   0.05 ,
we Do Not Reject H 0 .
Because we Do Not Reject H 0 , we are not finding evidence in support of the claim. Therefore, based on
our sample, it appears as though the proportion of people receiving parking tickets does not significantly
differ from p 25  0.20 , p25  0.80 as reported by the DMV.