Suppose the marketing research firm would like to examine if the social networki

Question

Suppose the marketing research firm would like to examine if the social networking site that a person primarily uses is influenced by his or her age. In a randomly drawn sample, 530 social network users were asked which site they primarily visited. These data are presented in the following table along with each person's age group:

At a significance level of 0.05 can we conclude that there is a relationship between age group of a social network user and the primary site they visit?

a. What is the question being asked?

b. What test should you run?

c. State the null and alternate hypotheses

d. Find the critical value

e. Compute the value of the test statistic

f. What is your decision regarding the null hypothesis? Interpret the result.

Social Media Age (years) Facebook Twitter LinkedIn 10-17 21 18 12 18-34 45 60 44 35-54 60 63 68 55 and older 61 43 35

Explanation / Answer

here there are two attributes: social media and age group where each age group is  divided into k=4 levels and social media is divided into l=3 levels

total number of observations=n=530

let fij denotes the number of persons for ith age group and jth social media      i=1,2,3,4   j=1,2,3

so from the table we have f11=21 f12=18   f13=12   f21=45 f22=60 f23=44 f31=60 f32=63 f33=68 f41=61 f42=43 f43=35

a) the question asked is : IS THE TWO ATTRIBUTES: SOCIAL MEDIA AND AGE GROUP INDEPENDENT OR NOT

b) the test to use is pearsonian chi square test for independence

c) null hypothesis: social media and age group are independent

alternative hypothesis: they are not independent

d) now under H0 the test statistic follows a chi square distribution with df (k-1)(l-1)

since here level of significance is alpha=0.05 hence the critical value is chi0.05;(k-1)(l-1) which is the upper alpha point of a chi square distribution with df (k-1)(l-1)

here k=4 and l=3   alpha=0.05 hence chi0.05;(4-1)(3-1)= 12.59159    [answer] [using MINITAB]

e) let f10=f11+f12+f13=21+18+12=51   f20=f21+f22+f23=45+60+44=149   f30=f31+f32+f33=60+63+68=191 f40=f41+f42+f43=61+43+35=139

f01=f11+f21+f31+f41=21+45+60+61=187   f02=f12+f22+f32+f42=18+60+63+43=184 f03=f13+f23+f33+f43=12+44+68+35=159

the test statistic is given by T=n[f11^2/(f10*f01)+f21^2/(f20*f01)+f31^2/(f30*f01)+f41^2/(f40*f01)+f12^2/(f10*f02)+f22^2/(f20*f02)+f32^2/(f30*f02)+f42^2/(f40*f02)+f13^2/(f10*f03)+f23^2/(f20*f03)+f33^2/(f30*f03)+f43^2/(f40*f03)-1]

hence the value of the test statistic is

t=530[21^2/(51*187)+45^2/(149*187)+60^2/(191*187)+61^2/(139*187)+18^2/(51*184)+60^2/(149*184)+63^2/(191*184)+43^2/(139*184)+12^2/(51*159)+44^2/(149*159)+68^2/(191*159)+35^2/(139*159)-1]=11.18072 [answer]

f) hence t<chi0.05;6 hence 5% level of significance H0 is accepted and the interpretation is that the two attributes are independent.

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