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A study of people who refused to answer survey questions provided the randomly selected sample data shown in the table below (based on data from 鈥淚 Hear You Knocking But You Can鈥檛 Come In,鈥 by Fitzgerald and Fuller, Sociological Methods and Research,Vol. 11, No. 1). At the 0.01 significance level, test the claim that the cooperation of

the subject (response or refusal) is independent of the age category. Does any particular age group appear to be particularly uncooperative?

Age


18-21

22-29

30-39

40-49

50-59

60 and over

Responded

73

255

245

136

138

202

Refused

11

20

33

16

27

49

Short Answer

Expert verified

The cooperation of the subject (response or refusal) is dependent on the age category.

The highest category of uncooperative subjects was in the age group of 60 and over.

Step by step solution

01

Given information

The data for the subject鈥檚 cooperation (response or refusal) and of the age category is provided.

02

Check the requirements of the test

Assume the subjects are randomly selected for the study

Use theformula for expected frequency as stated below,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The observed frequency table along with row and column total is represented as,


18-21

22-29

30-39

40-49

50-59

60 and over

Row total

Responded

73

255

245

136

138

202

1049

Refused

11

20

33

16

27

49

156

Column total

84

275

278

152

165

251

1205

Theexpected frequency tableis represented as,


18-21

22-29

30-39

40-49

50-59

60 and over

Responded

73.125

239.398

242.010

132.322

143.639

218.505

Refused

10.875

35.602

35.990

19.678

21.361

32.495

Here, all expected frequencies are greater than 5, which implies the requirements of the test are satisfied.

03

State the null and alternate hypothesis

The hypotheses are formulated as follows:

\({H_0}:\)The cooperation of the subject (response or refusal) is independent of age.

\({H_1}:\)The cooperation of the subject (response or refusal) is dependent of age.

04

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {73 - 73.125} \right)}^2}}}{{73.125}} + \frac{{{{\left( {255 - 239.398} \right)}^2}}}{{239.398}} + ... + \frac{{{{\left( {49 - 21.361} \right)}^2}}}{{21.361}}\\ = 20.271\end{aligned}\]

Therefore, the value of the test statistic is 20.271.

05

Compute the degrees of freedom

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {6 - 1} \right)\\ = 5\end{aligned}\)

Therefore, the degrees of freedom are 5.

06

Compute the critical value

From the chi-square table, the critical value for row corresponding to 5 degrees of freedom and at 0.01 level of significance 15.086.

The p-value is obtained as 0.0011.

07

State the decision

Since the critical value (15.086) is less than the value of the test statistic (20.271). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

08

State the conclusion

There is insufficient evidence to support claimthat the cooperation of the subject (response or refusal) is independent of age.

Thus, the cooperation of a subject is dependent on this age.

The proportions of subjects who were uncooperative in each category are stated below.

18-21

22-29

30-39

40-49

50-59

60 and over

Refused

0.1309\(\left( {\frac{{11}}{{84}}} \right)\)

0.0727

\(\left( {\frac{{20}}{{275}}} \right)\)

0.1187

\(\left( {\frac{{33}}{{278}}} \right)\)

0.1053

\(\left( {\frac{{16}}{{152}}} \right)\)

0.1636

\(\left( {\frac{{27}}{{165}}} \right)\)

0.1952

\(\left( {\frac{{49}}{{251}}} \right)\)

From the results, the most proportion of uncooperative subjects were in the age category 60 and over.

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Most popular questions from this chapter

In Exercises 5鈥20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

California Daily 4 Lottery The author recorded all digits selected in California鈥檚 Daily 4 Lottery for the 60 days preceding the time that this exercise was created. The frequencies of the digits from 0 through 9 are 21, 30, 31, 33, 19, 23, 21, 16, 24, and 22. Use a 0.05 significance level to test the claim of lottery officials that the digits are selected in a way that they are equally likely.

Using Yates鈥檚 Correction for Continuity The chi-square distribution is continuous, whereas the test statistic used in this section is discrete. Some statisticians use Yates鈥檚 correction for continuity in cells with an expected frequency of less than 10 or in all cells of a contingency table with two rows and two columns. With Yates鈥檚 correction, we replace

\(\sum \frac{{{{\left( {O - E} \right)}^2}}}{E}\)with \(\sum \frac{{{{\left( {\left| {O - E} \right| - 0.5} \right)}^2}}}{E}\)

Given the contingency table in Exercise 9 鈥淔our Quarters the Same as $1?鈥 find the value of the test \({\chi ^2}\)statistic using Yates鈥檚 correction in all cells. What effect does Yates鈥檚 correction have?

Motor Vehicle Fatalities The table below lists motor vehicle fatalities by day of the week for a recent year (based on data from the Insurance Institute for Highway Safety). Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Provide an explanation for the results.

Day

Sun.

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

Frequency

5304

4002

4082

4010

4268

5068

5985

Equivalent Tests A\({\chi ^2}\)test involving a 2\( \times \)2 table is equivalent to the test for the differencebetween two proportions, as described in Section 9-1. Using the claim and table inExercise 9 鈥淔our Quarters the Same as $1?鈥 verify that the\({\chi ^2}\)test statistic and the zteststatistic (found from the test of equality of two proportions) are related as follows:\({z^2}\)=\({\chi ^2}\).

Also show that the critical values have that same relationship.

Exercises 1鈥5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 鈥淏ody Data鈥 in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

Is the hypothesis test left-tailed, right-tailed, or two-tailed?

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