Analyze these data using the method presented in Chapter 8 for comparing two pro

Question

Analyze these data using the method presented in Chapter 8 for comparing two proportions (page 508). Analyze these data using the method presented in this chapter for examining a relationship between two categorical variables in 2 times 2 table. Use this example to explain the relationship between the chi-square test and the z test for comparing two proportions. The number of girls reported in this exercise is not the same as the number reported for Exercise 9.25. Suggest a possible reason for this difference. 9.28 Data for the boys. Refer to the previous exercise. Here are the corresponding data for boys: Using these data, repeat the analyses that you performed for the girls in Exercise 9.27. How do the results for the boys differ from those that you found lord girls?

Explanation / Answer

I don't know about which type of method you are suggesting in question 9.27(a) and 9,27(b) , but as shown in 9.27(c) , i can say it is Z - test and Chi - square test for poroprtions.

Q.9.27(a) Chi square test for proportiojns

Here Null Hypothesis H0 : p1 = p2

Alternative Hypothesis H1 : p1 p2

Now we have to calculate the expected value for each variable

I calculate following table by using proportions.

(a) Harassed online and in person = (361/1002) * 521 = 187.7

(b) not Harassed online but in person = (641/1002) * 521 = 333.3

(c) Harassed online but not in person = (361/1002) * 481 = 173.3

(d) not Harassed online and not in person = (641/1002) * 481 = 307.7

chi - square X2 = (321- 187.7)2/ 187.7 + (200- 333.3)2/ 333.3+ (40-173.3)2/ 173.3 + (441- 307.7)2/ 307.7

= 628.71

Degree of freedom here is 1

so by chi - square table p value = 5.2313 x 10 -69 ( i calculated it from Chitest in excel ) , which is too low that means that null hypothesis in invalid and both proportions are different.

Q . 9..28 (a) I am doing this part for Boys also , the chi square test

I calculate following table by using proportions.

Now similarly in Q/ 9.28, i have calculated the expected table bleow

chi - square X2 = (183- 80.8)2/ 80.8 + (154-256.2)2/ 256.2+ (48-150.2)2/ 150.2 + (578- 475.8)2/ 475.8

= 261.2945

Degree of freedom here is 1

so by chi - square table p value = 8.96079 x 10 -59 (i calculated it from Chitest in excel ) , which is too low that means that null hypothesis in invalid and both proportions are different.

Q. 9.28 (b) Now, we will solve it by Z -test

similarly,

Here Null Hypothesis H0 : p1 = p2

Alternative Hypothesis H1 : p1 p2

p1 (harassed in person) = 521/1002 = 0.52

p2(harassed online) = 361/1002 = 0.36

Z = (p1-p2) / sqrt [p*(1-p*) (1/n1+ 1/n2)

p* = (x1 + x2)/(n1 + n2 ) = ( 521 + 361)/ 2 * 1002 = 0.44

Z = (0.52 -0.36) / sqrt( 0.44 * 0.56 *2 /1002) = 7.214

so if for 95 % confidence interval, Z value must be of 1.96 so we can reject the null hypothesis and say both proportions are different.

SImilarly solving for males in Q. 9.28

similarly,

Here Null Hypothesis H0 : p1 = p2

Alternative Hypothesis H1 : p1 p2

p1 (harassed in person) = 337/963 = 0.35

p2(harassed online) = 231/963 = 0.24

Z = (p1-p2) / sqrt [p*(1-p*) (1/n1+ 1/n2)

p* = (x1 + x2)/(n1 + n2 ) = ( 337 + 231)/ 2 * 1002 = 0.295

Z = (0.35 -0.24) / sqrt( 0.295 * 0.705 *2 /1002) = 5.3989

so if for 95 % confidence interval, Z value must be of 1.96 so we can reject the null hypothesis and say both proportions are different for males.

(c) CHi- square test and Z - test gives same kind of result for both girls and boys.

(d) Not cmplete data given.

Harassed online Total Harassed in person Yes No Yes 321 200 521 No 40 441 481 Total 361 641 1002

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