Question

12. Boomerang Generation - Short Term: In a 2010 Pew Research Center survey, 86 out of 390 randomly selected young adults (ages 18-34) had to move back in with their parents after living alone. In a 2012 survey, 194 out of 808 young adults had to move back in with their parents. The table below summarizes this information. The standard error (SE) is given to save calculation time. total # who total # in proportion year moved back (x) survey (n) 2012 194 0.24010 2010 86 390 0.22051 Standard Error: SE = 0.02609 808 The Test: Test the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010. Test this claim at the 0.05 significance level. hinam

Answer 1

We want to test H0: p1 = p2 vs H1: p1 > p2

where p1 is the sample proportion in 2012 and p2 is that in 2010

The test statistic is

Z = ( p1 - p2 ) / [ p*(1-p)(1/n1 + 1/n2) ]^1/2

where p is the pooled proportion

If p1 = x1/n1 and p2 = x2/n2

then, p = (x1+x2) / (n1+n2)

The critical region at 0.05 significance level is Z > 1.645 (obtained from standard normal table )

Here p1 = 0.24010

p2 = 0.22051

n1 = 808

n2 = 390

So,on calculation, Z = 0.7508

Z < 1.645

So, we fail to reject Ho.

So, there is not enough evidence to claim that a greater proportion of young adults moved back in with their parents in 2012 than in 2010