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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.

Educational attainment 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 education attainments 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= educational attainments 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 educational attainment and response.
Alternative hypothesis:
H1 : There is an association exists between educational attainment 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: Postgraduate, College graduate, Some college, No college

From the MINITAB output,the value of the chi-square statistic is13.651

06

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

From the MINITAB output, the P-value is =0.034

Rejection rule:
If P-value≤α, then reject the null hypothesis.
Here, theP-value is lesser than the level of significance,
That is, P-value0.034<α=(0.005).
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 education attainments and response at the 5% significance level.

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Most popular questions from this chapter

In each case, decide whether Assumptions 1and 2for using chi-square goodness-of-fit test are satisfied.

Sample size:n=50.

Relative frequencies:0.20,0.20,0.25,0.30,0.05.

In each of the given Exercises, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, x, and y, of a population. For each exercise, perform the following tasks.

a. Find the expected frequencies. Note: You will first need to compute the row totals, column totals, and grand total.

b. Determine the value of the chi-square statistic.

c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated.

In each of the given Exercises, we have given the number of possible values for two variables of a population. For each exercise, determine the maximum number of expected frequencies that can be less than 5 in order that Assumption 2 of Procedure 12.2 on page 506 to be satisfied. Note: The number of cells for a contingency table with m rows and n columns is mâ‹…n.

12.72 four and three

If a variable of two populations has only two possible values, the chi-square homogeneity test is equivalent to a two-tailed test that we discussed in an earlier chapter. What test is that?

Loaded Die? A gambler thinks a die may be loaded, that is, that the six numbers are not equally likely. To test his suspicion, he rolled the die 150 times and obtained the data shown in the following table.

Number

1

2

3

4

5

6

Frequency

23

26

23

21

31

26


Do the data provide sufficient evidence to conclude that the die is loaded? Perform the hypothesis test at the 0.05 level of significance.
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