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Job Satisfaction. A CNN/USA TODAY poll conducted by Gallul asked a sample of employed Americans the following question: "Which do you enjoy more, the hours when you are on your job, or the hours when you are not on your job?" The responses to this question were cross-tabulated against several characteristics, among which were gender, age, type of community, educational attainment, income, and type of employer. The data are provided on the WeissStats site. In each of Exercises 12.87-12.92, use the technology of your choice to decide, at the 5% significance level, whether an association exists between the specified pair of variables.

type of employer and response

Short Answer

Expert verified

The null hypothesis is rejected at 5% level.
The results are statistically significant at 5% level of significance.

There is an association exists between type of employer and response at the 5% significance level.

Step by step solution

01

Step 1. Given information

The given significance level =5%

The given specified pair of variables= type of employer and response

02

Step 2. Check whether or not there is association exists between age and response at 5%significance level.

Step 1:
The test hypotheses are given below:
Null hypothesis:
H0 : There is no association exists between type of employer and response.
Alternative hypothesis:
H1 : There is an association exists between type of employer and response.

03

Step 3.  Finding the level of significance

Step 2: Decide the level of significance.
Here, the level of significance is,α=0.05

04

Step 4. Find the expected frequency and test statistic. 

MINITAB procedure:
Step 1: Choose Stat>Tables>Chi-Square Test (Two-Way Table in Worksheet).
Step 2: In Columns containing the table, enter the columns of Postgraduate, College graduate, Some college and No college.
Step 3: Click OK.

05

Step 5. Finding the MINITAB output

Chi-Square Test for Association: Private ,governament and self

From the MINITAB output,the value of the chi-square statistic is49.665

06

Step 6. Finding p-value and check the solution by rejection and interpretation 

From the MINITAB output, the P-value is 0.000

Rejection rule:
If P-value≤α, then reject the null hypothesis.
Here, theP-value is lesser than the level of significance,
That is,P-value (=0.000)<α(=0.05).
Therefore, the null hypothesis is rejected at5%level.
Thus, the results are statistically significant at 5% level of significance.

Interpretation:
Thus, there is an association exists between type of employer and response at the 5% significance level.

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Most popular questions from this chapter

In each of Exercises 12.18-12.23, we have provided a distribution and the observed frequencies of the values of a variable from a simple random sample of a population. In each case, use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution.

Distribution: 0.2, 0.4, 0.3, 0.1

Observed frequencies: 85, 215, 130, 70

Significance level = 0.05

For a χ2curve with 22degrees of freedom, determine the χ2value that has area

a. 0.01to its right.

b. 0.005to its right.

Identify three techniques that can be tried as a remedy when one or more of the expected-frequency assumptions for a chi-square independence test are violated.

Job Satisfaction. A CNN/USA TODAY poll conducted by Gallul asked a sample of employed Americans the following question: "Which do you enjoy more, the hours when you are on your job, or the hours when you are not on your job?" The responses to this question were cross-tabulated against several characteristics, among which were gender, age, type of community, educational attainment, income, and type of employer. The data are provided on the WeissStats site. In each of Exercises 12.87-12.92, use the technology of your choice to decide, at the 5% significance level, whether an association exists between the specified pair of variables.

Educational attainment and response

In each of Exercises 12.18-12.23, we have provided a distribution and the observed frequencies of the values of a variable from a simple random sample of a population. In each case, use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution.

Distribution: 0.2, 0.4, 0.3, 0.1

Observed frequencies: 39, 78, 64, 19

Significance level = 0.05

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