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4. Facebook, Instagram, and LinkedIn are social networking sites on the Internet. In 2009, the Pew...

Question

4. Facebook, Instagram, and LinkedIn are social networking sites on the Internet. In 2009, the Pew...

4. Facebook, Instagram, and LinkedIn are social networking sites on the Internet. In 2009, the Pew Internet and American Life
4. Facebook, Instagram, and LinkedIn are social networking sites on the Internet. In 2009, the Pew Internet and American Life Project found evidence that the mean ages for the users of these three sites were not equal. Suppose that random samples of Facebook users, MySpace users, and LinkedIn users were taken this year and their ages are recorded in the following table. Assume all three distributions are approximately normal. Facebook Instagram Linkedin 25 34 28 25 24 30 2249 a) What is the response variable? b) What is the factor and how many levels does the factor have? (2) c) Is there evidence that the population mean age of the users users of social media differs for at least two groups? Conduct the appropriate test at the 5% level of significance. (2)Define parameter (4)Null and alternative hypothesis (2)test statistic (Symbol and value) (3)P value (4)Conclusion Place ANOVA table here b) Perform the appropriate Tukey multiple comparisons using a family confidence level of 0.95. State your conclusion using clear, complete sentences. Be sure your conclusion clearly states which social networking sites have significantly different means and by how much. (8) Place Tukey multiple comparisons here

Answers

response variable is number of uesers

factor is social networking site and 3 levels are there

....................

A B C
count, ni = 7 7 7
mean , x̅ i = 26.000 27.00 40.00
std. dev., si = 5.292 6.028 6.000
sample variances, si^2 = 28.000 36.333 36.000
total sum 182 189 280 651 (grand sum)
grand mean , x̅̅ = Σni*x̅i/Σni =   31.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² 25.000 16.000 81.000
TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² = 175.000 112.000 567.000 854
SS(within ) = SSW = Σ(n-1)s² = 168.000 218.000 216.000 602.0000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   21
df within = N-k =   18
  
mean square between groups , MSB = SSB/k-1 =    427.0000
  
mean square within groups , MSW = SSW/N-k =    33.4444
  
F-stat = MSB/MSW =    12.7674
P value =   0.0004

.anova table
SS df MS F p-value F-critical
Between: 854.00 2 427.00 12.77 0.0004 3.55
Within: 602.00 18 33.44
Total: 1456.00 20
α = 0.05
conclusion : p-value<α , reject null hypothesis    

p value <0.05, reject Ho

means are different

.........................

Level of significance 0.05
no. of treatments,k 3
DF error =N-k= 18
MSE 33.444
q-statistic value(α,k,N-k) 3.6100
critical value = q*√(MSE/n)
confidence interval
population mean difference critical value lower limit upper limit result
µ1-µ2 -1.00 7.89 -8.89 6.89 means are not different
µ1-µ3 -14.00 7.89 -21.89 -6.11 means are different
µ2-µ3 -13.00 7.89 -20.89 -5.11 means are different

.....................

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