Elementary Statistics: A Step By Step Approach
Elementary Statistics: A Step By Step Approach
10th Edition
ISBN: 9781259755330
Author: Allan G. Bluman
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 9, Problem 9.1.1RE

For each exercise, perform these steps. Assume that all variables are normally or approximately normally distributed.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

Section 9–1

1. Driving for Pleasure Two groups of randomly selected drivers are surveyed to see how many miles per week they drive for pleasure trips. The data are shown. At α = 0.01, can it be concluded that single drivers do more driving for pleasure trips on average than married drivers? Assume σ1 = 16.7 and σ2 = 16.1.

Chapter 9, Problem 9.1.1RE, For each exercise, perform these steps. Assume that all variables are normally or approximately

a.

Expert Solution
Check Mark
To determine

To identify: The claim and state   H0 and H1

Answer to Problem 9.1.1RE

The claim is that “single drivers do more driving for pleasure trips on average than married drivers”.

Null hypothesis: H0:μ1=μ2

Alternative hypothesis: H1:μ1>μ2 (claim)

Explanation of Solution

Given info:

Two groups of randomly selected drivers are surveyed to see how many miles per week they drive for pleasure trips.

Justification:

Here, the claim is that “single drivers do more driving for pleasure trips on average than married drivers”. This can be written as μ1>μ2 . The complement of the claim is μ1=μ2 . In the given experiment, the alternative hypothesis indicates the claim.

b.

Expert Solution
Check Mark
To determine

To find: The critical-value.

Answer to Problem 9.1.1RE

The critical -value is 2.326.

Explanation of Solution

Calculation:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability> OK.
  • From Distribution, choose ‘Normal’ distribution.
  • Click the Shaded Area tab.
  • Choose ProbabilityValue and Both tail for the region of the curve to shade.
  • Enter the Probability value as 0.01.
  • Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9, Problem 9.1.1RE , additional homework tip  1

Thus, the critical value for right tail test is 2.326.

The rejection region for a two-tailed test with α=0.01 is, z>+z0(=+2.326) .

c.

Expert Solution
Check Mark
To determine

To find: The standardized test statistic z.

Answer to Problem 9.1.1RE

The standardized test statistic z is 0.59.

Explanation of Solution

Calculation:

Software procedure:

Step by step procedure to obtain the mean using the MINITAB software:

  • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • In Variables enter the columns Single drivers and Married drivers.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 9, Problem 9.1.1RE , additional homework tip  2

From the output, the mean for single drivers is 120.09 and the mean for married drivers is 117.80.

Test statistic:

z=(X¯1X¯2)(μ1μ2)σ12n1+σ22n1=(120.09117.80)(16.7)235+(16.1)235=2.293.9=0.59

d.

Expert Solution
Check Mark
To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of α=0.01 .

Answer to Problem 9.1.1RE

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Justification:

Decision rule:

  • If z>z0(=+2.326) , then reject the null hypothesis.

Here, the test statistic is less than the critical value. That is, z(=0.59)<z0(=+2.326) .

Thus, the decision is “fail to reject the null hypothesis”.

e.

Expert Solution
Check Mark
To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 9.1.1RE

The conclusion is thatthere is no enough evidence to support the claim that single drivers do more driving for pleasure trips on average than married drivers.

Explanation of Solution

From part (d), the null hypothesis is not rejected. Thus, there is no enough evidence to support the claim that single drivers do more driving for pleasure trips on average than married driversat 1% level of significance.

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Chapter 9 Solutions

Elementary Statistics: A Step By Step Approach

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