Homework 9 Chapter 15 Your Name

  • Uncategorized

Homework9: Chapter 15

YourName

11.Inthe general population, political party affiliation is 30%Republican, 55% Democratic, and 15% other. To determine whether thisdistribution is also found among the elderly, in a sample of 100senior citizens, we find 18 Republicans, 64 Democrats, and 18 other.

(a)What procedure should we perform?

Thebest procedure to perform in this question is to conduct a one-waychi square procedure since the data given has frequency of categorymembership along each variable.

(b)What are H0and Ha?

Itis claimed from the data to find out whether the affiliation to apolitical party is as follows

Republican– 30%

Democrats– 55%

Other– 15%

Withthis claim the null and alternative hypotheses will be:

Ho:The elderly affiliation to a political party is distribted with 30%Republican, 55% Democrats and 15% other

Ha:The elderly political party affiliation is not distributed in thisway

(c)What is fefor each group?

Weshall calculate fe for each of the three groups. To get fe wemultiply the percentage of the particular group with the total numberof senior citizens

Fevalue for the democrats therefore = 0.55 * 100 = 55

Fevale for the republicans = 0.3*100 = 30

Fefor others = 0.15*100 = 15

(d)Compute X2obt.

Wherefo is the observed frequency and fe represents the expected frequency

ThereforeX2obt=(18-30)2/30+(64 – 55)2/55+(18-15)2/15

=144/30 +81/55 +9/15

= 4.8 + 1.47 + 0.6 = 6.87

(e)With α = .05, what do you conclude about party affiliation amongsenior citizens?

Beforewe can conclde, we have to compare the chi square obtained orcalculated with the value taken from the chi square tables

Sincewe are performing a one-way chi square procedure, the degrees offreedom is given by subtracting one from the number of categories. Inthis case

Df= 3 -1 = 2

Fromthe tables, a chi sqare value with 0.05 significance level and 2degrees of freedom is 5.9915

Itis clear that the obtained chi square value (6.87) is bigger than thecritical value of 5.9915 at 0.05 level of significance. In this casewe therefore reject the null hypothesis thus concluding that theaffiliation of the elderly to the political party is not distributedwith 30% Republican, 55% Democrats and 15% other

14.Thefollowing data reflect the frequency with which people voted in thelast election and were satisfied with the officials elected:

Satisfied

YesNo

48

35

33

52

Yes

Voted

No

  1. What procedure should we perform?

Inthis case, we have data with frequency of category membership alongtwo variable which suggests a two-way chi-square test

  1. What are H0 and Ha?

Ho:People voted in the election and their satisfaction with the electedofficials is independent

Ha:People voted in the election and their satisfaction with electedofficials is dependent

thishypothesis will be tested at 0.05 level of significance

  1. What is fe in each cell?

Fora two way chi square test, the expected frequency is got by dividingthe product of the total observed frequency of the row and that ofthe column and the total number

i.efe = (cell’s row total)(cell’s column total)/N

thereforewe have these total

forthose who voted yes row total = 48 + 35 = 83

forthose who voted No row total = 33+52 = 85

forcolmn total we have

SatisfiedYes = 48+33 = 81

SatisfiedNo = 35 + 52 = 87

TotalN = 83+85 = 81+87 = 168

Thereforefe for the first cell (voted yes Satisfied yes) = 83*81/168 = 40.018

Fefor the second cell (Voted Yes Satisfied No) = 83*87/168 = 42.985

Fefor the third cell (Voted No, satisfied Yes) = 85*81/168 = 40.982

Fefor the last cell (voted No, satisfied No) = 85*87/168 = 44.018this is summarized as follows

Yes

No

Row total

Yes

Fe = 40.018

Fe=42.982

83

No

Fe=40.982

Fe=44.018

85

Column total

81

87

N=168

  1. Compute X 2obt.

=(48-40.018)2/40.018+ (35-42.982)2/42.982+(33-40.982)2/40.982+(52 -44.018)2/44.018

  1. With α = .05, what do you conclude about the correlation between these variables?

Firstof all we calculate the degrees of freedom which is given as

Df= (nmber of rows – 1)(number of columns – 1) = (2 – 1) (2 –1) = 1

Readingthe critical value of the chi square with 1 degree of freedom and0.05 significant level we get 3.842

Itis therefore clear that the calculated value of the chi square =6.076 is greater than the tabulated value of the chi square at 1degree of freedom and 0.05 level of significance = 3.842. Thereforewe have the evidence to allege that the null hypothesis is not truethus we conclude that the two variables the people voted in theelection and their satisfaction of the elected officials aredependent.

(f)How consistent is this relationship?

Tocalculate the consistence of this relationship, we find the squareroot of obtained chi square divided by the total number (N)and theobtained chi square

Mathematically,C =

= = 0.2159

Thisindicates that the relationship is consistent

15.Astudy determines the frequency of the different political partyaffiliations for male and female senior citizens. The following dataare obtained:

RepublicanDemocrat Other

18

43

14

39

23

18

Male

Gender

Female

  1. What procedure should we perform?

Atwo-way chi square procedure is appropriate for this since themembership of the category is along two variables

  1. What are H0 and Ha?

Ho:gender and political party affiliation are independent in thepopulation

Ha:gender and political party affiliation are not independent in thepopulation

  1. What is fe in each cell?

Columntotal for republican = 57, for Democrat = 66, for other = 32

Rowtotal for male = 75, for female = 80

TotalN = 155

Fefor each cell is as follows

Forthe cell (Male, Republican) = 75*57/155 = 27.581

Cell(female, Republican) = 80*57/155 = 29.419

Cell(male, Democrat) = 75*66/155 = 31.935

Cell(female, Democrat) = 80*66/155 = 34.065

Cell(Male, Other) = 75*32/155 = 15.484

Cell(female, other) = 80*32/155 = 16.516

Ina table summary, we have the cell fe’s as

Republican

Democrat

Other

Row total

Male

27.581

31.935

15.484

75

Female

29.419

34.065

16.516

80

Colmn total

57

66

32

155

  1. Compute X2obt.

X2obt= (18 – 27.581)2/27.581+ (43 – 31.935)2/31.936+ (14 – 15.484)2/15.484+(39 – 29.419)2/29.419+ ( 23 – 34.065)2/34.065+ (18 – 16.516)2/16.516

=3.328 + 3.834 + 0.142 +3.12 +3.594 +0.133

=14.152

  1. With α = .05, what do you conclude about gender and party affiliation in the population of senior citizens?

Toconclude we must get the critical value of the chi square from thechi square tables with degrees of freedom = (2 – 1) ( 3 – 1) = 2and 0.05 level of significance

Fromthe tables, the critical value = 5.9915

Sincethe calculated value of the chi square (14.152) exceeds the criticalvalue at 0.05 significant level with 2 degrees of freedom (5.9915),we reject the null hypothesis and conclude that the gender andpolitical party affiliation are dependent in the population.

  1. How consistent is this relationship?

Consistency= = 0.289 suggesting that the consistence of the relationship

17.Aftertesting 40 participants, a significant X2obt of 13.31 was obtained.With α = .05 and df=2, how would this result be reported in a publication?

Wehave the total number of participants N= 40

Thecalculated chi square value = X2obt= 13.31

Levelof significance = 0.05 and degrees of freedom = 2

Wefirst of all state the null and alternative hypotheses which are

Ho:The variables presented are independent

Ha:The variables presented are not independent

Wetherefore go ahead and find the tabulated chi square value with 2degrees of freedom and 0.05 level of significance which is thep-value

Fromthe table we get that X2crit= 5.9915

Inconclusion as we can see, X2obt= 13.31 is larger than X2crit= 5.9915, a clear indication that the results are significant. Thuswe refuse to accept the null hypothesis and conclude that thevariables are not independent.

Ina report form this is reported as X2(2, N =40) = 13.31, p&lt0.05

Close Menu