Bluman, Elementary Statistics: A Step By Step Approach, © 2015, 9e, Student Edition (reinforced Binding) (a/p Statistics)
Bluman, Elementary Statistics: A Step By Step Approach, © 2015, 9e, Student Edition (reinforced Binding) (a/p Statistics)
9th Edition
ISBN: 9780021418251
Author: Allan G. Bluman
Publisher: MCGRAW-HILL HIGHER EDUCATION
Question
Book Icon
Chapter 13.6, Problem 24E
To determine

To find : The outcomes of the coin by tossing it 30 times.

To test: Whether the sequence of the outcome of the coin is random.

To repeat: The experiment and test for randomness.

To compare : Both the results.

Expert Solution & Answer
Check Mark

Answer to Problem 24E

There is sufficient evidence to infer that the sequence of the outcome of the coin is random is random.

In both the experiments the outcome of the coin occurs at random.

Explanation of Solution

Given info:

The task is to toss a coin 30 times and note the results and test for randomness. The level of significance is α=0.05 .

Calculation:

First experiment:

Outcomes of the coin:

Software Procedure:

Step by step procedure to generate outcomes of coin toss using the MINITAB software:

  • Choose Calc > Random Data > Data from columns.
  • Select the column Random.
  • Enter 30 under Number of rows of data to generate.
  • In Store column Choose Outcome.
  • Click OK.

Output using the MINITAB software is given below:

Bluman, Elementary Statistics: A Step By Step Approach, © 2015, 9e, Student Edition (reinforced Binding) (a/p Statistics), Chapter 13.6, Problem 24E , additional homework tip  1

The testing hypotheses are given below:

Null hypothesis:

 H0: The sequence of the outcome of the coin is random.

Alternative hypothesis:

H1: The sequence of the outcome of the coin is not random.

Critical value:

Here, the test is two tailed test.

For the level of significance α=0.05 ,

From the data it can be seen that there are 16 H’s and 14 T’s.

From Table M of the Number of Runs, the critical value that corresponds to α=0.05 and n1=16,n2=14 is 10 and 22.

Thus, the critical value for the number of runs is 10 and 22.

Test statistic:

The number of runs for the given sequence is obtained as follows:

The list of the runs is,

Run Letters
1 T,T
2 H
3 T
4 H
5 T,T
6 H,H
7 T
8 H
9 T,T
10 H,H
11 T
12 H
13 T,T,T
14 H,H,H,H,H
15 T
16 H,H,H
17 T

Here, the number of runs is 17.

Thus the test value is G=17 .

Decision rule for a two tailed test:

  • If the test value is does not lies between the critical value, then reject the null hypothesis H0 .

Conclusion:

The value of test statistic is 17 and the critical value is 10 and 22.

Here, the test statistic value lies between the critical value.

That is, 10(=criticalvalue)<17(=G)<22(=criticalvalue)

Thus, the decision is “not to reject the null hypotheses”.

Thus, it can be concluded that there is enough evidence to support the claim that the sequence of the outcome of the coin is random.

Second experiment:

Outcomes of the coin:

Software Procedure:

Step by step procedure to generate outcomes of coin toss using the MINITAB software:

  • Choose Calc > Random Data > Data from columns.
  • Select the column Random.
  • Enter 30 under Number of rows of data to generate.
  • In Store column Choose Outcome.
  • Select sample with replacement.
  • Click OK.

Output using the MINITAB software is given below:

Bluman, Elementary Statistics: A Step By Step Approach, © 2015, 9e, Student Edition (reinforced Binding) (a/p Statistics), Chapter 13.6, Problem 24E , additional homework tip  2

The testing hypotheses are given below:

Null hypothesis:

 H0: The sequence of the outcome of the coin is random.

Alternative hypothesis:

H1: The sequence of the outcome of the coin is not random.

Critical value:

Here, the test is two tailed test.

For the level of significance α=0.05 ,

From the data it can be seen that there are 16 H’s and 14 T’s.

From Table M of the Number of Runs, the critical value that corresponds to α=0.05 and n1=16,n2=14 is 10 and 22.

Thus, the critical value for the number of runs is 10 and 22.

Test statistic:

The number of runs for the given sequence is obtained as follows:

The list of the runs is,

Run Letters
1 T,T
2 H
3 T
4 H
5 T,T
6 H,H
7 T
8 H
9 T,T
10 H,H
11 T
12 H
13 T,T,T
14 H,H,H,H,H
15 T
16 H,H,H
17 T

Here, the number of runs is 17.

Thus the test value is G=17 .

Decision rule for a two tailed test:

  • If the test value is does not lies between the critical value, then reject the null hypothesis H0 .

Conclusion:

The value of test statistic is 17 and the critical value is 10 and 22.

Here, the test statistic value lies between the critical value.

That is, 10(=criticalvalue)<17(=G)<22(=criticalvalue)

Thus, the decision is “not to reject the null hypotheses”.

Thus, it can be concluded that there is enough evidence to support the claim that the sequence of the outcome of the coin is random.

Comparison:

Here, the results in both the experiment is same.

From the test, it can be concluded that the outcome of the coin occurs at random.

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Chapter 13 Solutions

Bluman, Elementary Statistics: A Step By Step Approach, © 2015, 9e, Student Edition (reinforced Binding) (a/p Statistics)

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