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-valueα, 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 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.

A chi-square homogeneity test is to be conducted to decide whether a difference exists among the distributions of a variable of six populations. The variable has five possible values. What are the degrees of freedom for the x2-statistic?

In Exercises 12.101-12.106, tase either the critical-value approach or the P-value approach to perform a chi-square homogeneity test, provided the conditions for using the test are met.
12.101 Self-Concept and Sightedness. Self-concept can be defined as the general view of oneself in terms of personal value and capabilities. A study of whether visual impairment affects self-concept was reported in the article "An Exploration into Self Concept: A Comrarative Analysis between the Adolescents Who Are Sighted and
Elind in India" (British Journal of Visual Impairment, Vol. 30, No. 1, of sighted and blind Indian adolescents gave the following data on self-concept.

a. At the 5% significance level, do the data provide sufficient evidence to conclude that a difference exists in self-concept distributions between sighted and blind Indian adolescents?
b. Repeat part (a) at the1%significance level.

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.

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.

Education of Prisoners. In the article "Education and Correctional Populations" (Bureau of Justice Statistics Special Report, NCJ 195670), C. Harlow examined the educational attainment of prisoners by type of prison facility. The following contingency table was adapted from Table 1 of the article. Frequencies are in thousands, rounded to the nearest hundred.


StateFederalLocalTotal
8th grade or less149.910.666.0226.5
some high school269.112.9168.2450.2
GED300.820.171.0391.9
High school diploma216.424.0130.4370.8
Postsecondary95.014.051.9160.9
College grad and more25.37.216.148.6
Total1056.588.8503.61648.9

How many prisoners

a. are in state facilities?

b. have at least a college education?

c. are in federal facilities and have at most an 8th-grade education?

d. are in federal facilities or have at most an 8th-grade education? e. in local facilities have a postsecondary educational attainment

f. who have a postsecondary educational attainment are in local facilities?

g. are not in federal facilities?

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