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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-valueα, 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 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 m⋅n.

12.69 four and five

In each of Exercises 12.11-12.16, we have given the relative frequencies for the null hypothesis of a chi-square goodness-of-fir text and the sample size. In each case, decide whether Assumptions 1 and 2 for using that text are satisfied.

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Relative frequencies: 0.44 , 0.25 , 0.30 , 0.01.

12.51 Farms. The U.S. Department of Agriculture publishes information about U.S. farms in Census of Agriculture. A joint frequency distribution for number of farms, by acreage and tenure of operator, is provided in the following contingency table. Frequencies are in


Full ownerPart ownerTenantTotal
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368
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a. Fill in the six missing entries.

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e. How many farms are operated by part owners and have between 500 acres and 999 acres, inclusive?

f. How many farms are not full-owner operated?

g. How many tenant-operated farms have 180 acres or more?

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At the 1%significance level, do the data provide sufficient evidence to conclude that a difference exists in race distributions among the four U.S. regions?

Scoliosis is a condition involving curvature of the spine. In a study by A. Nachemson and L. Peterson, reported in the Journal of Bone and Joing Surgery, 286girls aged 10to 15years were followed to determine the effects of observation only (129patients), an underarm plastic brace (111patients), and nighttime surface electrical stimulation (46 patients). A treatment was deemed to have failed if the curvature of the spine increased by 6 on two successive examinations. The following table summarizes the results obtained by the researchers.

At the 5%significance level, do the data provide sufficient evidence to conclude that a difference in failure rate exists among the three types of treatments?

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