Question

The term "Boomerang Generation" refers to the recent generation of young adults (ages 18-34) who have had to move back in with their parents. In a 2018 national survey, 210 out of 840 randomly selected young adults had to move back in with their parents. In a similar 2008 national survey, 150 out of 800 randomly selected young adults had to move back in with their parents. You would like to determine if there is a significant difference in the proportion of young adults who moved back in with their parents in 2018 and 2008?

(a) State the research question.

Is there a significant difference in the proportion of young adults who moved back in with their parents in 2018 and 2008?

(b) State both the null and alternative hypotheses.

(c) Calculate the standard error.

(d) Calculate the test statistic, and state the decision regarding the null hypothesis.

(e) Calculate the p-value for the test.

(f) Assess the strength of your decision in part (d) using the p-value.

(g) Answer the research question. How would you describe the current trend?

(h) Calculate a 95% confidence interval for the difference between the two corresponding population proportions (see pages 217-218 in the Blue Book).

(i) How does your result in part (h) relate to the decision regarding the null hypothesis in parts (d), (e), and (f)? Be specific.

Answer 1

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0         
                  
sample #1   ----->   experimental          
first sample size,     n1=   840          
number of successes, sample 1 =     x1=   210          
proportion success of sample 1 , p̂1=   x1/n1=   0.2500          
                  
sample #2   ----->   standard          
second sample size,     n2 =    800          
number of successes, sample 2 =     x2 =    150          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.188          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2500   -   0.1875   =   0.0625
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.2195          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0204      

   
Z-statistic = (p̂1 - p̂2)/SE = (   0.063   /   0.0204   ) =   3.06
                  
  
p-value =        0.0022   [excel formula =2*NORMSDIST(z)]      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion:   There is enough evidence to SAY THAT TWO PROPORTIONS ARE DIFFERENT

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

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0203          
margin of error , E = Z*SE =    1.960   *   0.0203   =   0.0399
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.063   -   0.0399   =   0.0226
upper limit = (p̂1 - p̂2) + E =    0.063   +   0.0399   =   0.1024
                  
so, confidence interval is (   0.0226   < p1 - p2 <   0.1024   )  
..........

CI does not conatin 0 , so reject HO, RESULT SIGNIFICANT

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Please revert back in case of any doubt.

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