ELEMENTARY STATISTICS W/CONNECT >IP<
ELEMENTARY STATISTICS W/CONNECT >IP<
4th Edition
ISBN: 9781259746826
Author: Bluman
Publisher: MCG
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Chapter 13.6, Problem 28E
To determine

To test: Whether the sequence of the people who were tested for drivers wear glasses or contact lenses is random.

Expert Solution & Answer
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Answer to Problem 28E

There is sufficient evidence to infer that the sequence of the people who were tested for drivers wear glasses or contact lenses is random.

Explanation of Solution

Given info:

The data represents the sequence of the people who were tested for drivers who wear glasses (G) or contact lenses(C). The level of significance is α=0.05 .

Calculation:

The testing hypotheses are given below:

Null hypothesis:

 H0: The sequence of the people who were tested for drivers is random.

Alternative hypothesis:

H1: The sequence of the people who were tested for drivers 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 31 C’s and 19 G’s.

That is, the sizes are n1=31 and n2=19 .

From Table E, The Standard Normal Distribution, the critical value that corresponds to the level of significance α=0.05 and ±1.96 .

Thus, the critical value is ±1.96 .

Test statistic:

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

The list of the runs is,

Run Letters
1 G
2 C,C,C
3 G
4 C,C
5 G,G
6 C,C,C
7 G,G
8 C
9 G
10 C,C
11 G,G
12 C,C,C
13 G,G
14 C,C
15 G
16 C
17 G
18 C,C,C,C
19 G
20 C
21 G,G,G,G
22 C,C,C,C,C,C,C,C,C
23 G

Here, the number of runs is G=23 .

The mean number of runs is,

μG=2n1n2n1+n2+1=2×31×1931+19+1=1,17850+1=24.56

The standard deviation of runs is,

σG=2n1n2(2n1n2n1n2)(n1+n2)2(n1+n21)=(2×31×19)((2×31×19)3119)(31+19)2(31+191)=1,178×1,1282,500×49=1,328,784122,500

     =10.8472=3.2935

The test statistic value is,

z=GμGσG=2324.563.2935=1.563.2935=0.4736

Hence, the test value is z=0.4736

Decision rule:

  • If the negative test value is less than the negative critical value, then reject the null hypothesis H0 .
  • If the negative test value is greater than the negative critical value, then fail to reject the null hypothesis H0 .

Conclusion:

The test value is, z=0.4736 and the critical value is -1.96.

Here, the negative test statistic value is greater than the negative critical value.

That is, 0.4736(=z)>1.96(=criticalvalue)

By the rejection rule “failed to reject the null hypotheses”.

Thus, it can be concluded that there is enough evidence to support the claim that the sequence of the people who were tested for drivers wear glasses or contact lenses is random.

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Chapter 13 Solutions

ELEMENTARY STATISTICS W/CONNECT >IP<

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