Solution Found!
Refer to the following list of numbers of
Chapter 12, Problem 10 CRE(choose chapter or problem)
Please be aware that some of the following problems may require knowledge of concepts presented in previous chapters.
In Exercises 15, refer to the following list of numbers of years that U.S. presidents, popes, and British monarchs lived after their inauguration, election, or coronation, respectively. (As of this writing, the last president is Gerald Ford, the last pope is John Paul II, and the last British monarch is George VI.) Assume that the data are samples randomly selected from larger populations.
Lottery: GoodnessofFit The bars in the histogram included with Exercise 9 depict these frequencies: 21, 19, 15, 18, 24, 18, 16, 24, 30, and 15.
Test the claim that the digits are selected from a population in which the digits are all equally likely. Is there a problem with the lottery?
Questions & Answers
QUESTION:
Please be aware that some of the following problems may require knowledge of concepts presented in previous chapters.
In Exercises 15, refer to the following list of numbers of years that U.S. presidents, popes, and British monarchs lived after their inauguration, election, or coronation, respectively. (As of this writing, the last president is Gerald Ford, the last pope is John Paul II, and the last British monarch is George VI.) Assume that the data are samples randomly selected from larger populations.
Lottery: GoodnessofFit The bars in the histogram included with Exercise 9 depict these frequencies: 21, 19, 15, 18, 24, 18, 16, 24, 30, and 15.
Test the claim that the digits are selected from a population in which the digits are all equally likely. Is there a problem with the lottery?
ANSWER:
Answer:
Step 1 of 2
The hypotheses can be written as
H0 : The digits are selected from a population in which the digits are all equally likely
H1 : At least one of the digits selected from a population in which the digits are not equally likely.
Use a 0.05 significance level
SL. No. |
O |
p = 1/10 |
E = np |
(O - E )2 |
(O - E)2/E |
1 |
21 |
0.1 |
20 |
1 |
0.05 |
2 |
19 |
0.1 |
20 |
1 |
0.05 |
3 |
15 |
0.1 |
20 |
25 |
1.25 |
4 |
18 |
0.1 |
20 |
4 |
0.2 |
5 |
24 |
0.1 |
20 |
16 |
0.8 |
6 |
18 |
0.1 |
20 |
4 |
0.2 |
7 |
16 |
0.1 |
20 |
16 |
0.8 |
8 |
24 |
0.1 |
20 |
16 |
0.8 |
9 |
30 |
0.1 |
20 |
100 |
5 |
10 |
15 |
0.1 |
20 |
25 |
1.25 |
Sum |
200 |
1 |
200 |
208 |
10.4 |
Expected frequency (E) = np
= 200(1/10)
= 20
The calculation continues as follows. Letting E be the expected frequency of an outcome and O be the observed frequency of that outcome.