Living at Home:

According to a 2011 report from the U.S. Census, 59% of young men (ages 18-24) are living at home with their parents. Is the percentage higher in your community?

Suppose we select a random sample of 20 young men from your community. We cannot use a normal model for the sampling distribution because one of the conditions is not met. We expect 41% of the young men in the sample are not living at home, so [tex]n(1–p) = 20(0.41) = 8.2[/tex], which is less than 10.

We ran a simulation with [tex]p = 0.59[/tex] and [tex]n = 20[/tex]. Suppose that the sample of 20 young men from your community has 14 living at home. This is 70% of the sample. In the simulated sampling distribution, 446 out of 2000 samples have a sample proportion of 0.70 or higher.

Do the data suggest that the percentage of young men living at home in your community is higher than 59%?

The data does not have strong enough evidence to conclusively say that the percentage of young men living at home in your community is significantly higher than the national percentage of 59%.

Check the conditions for a normal approximation:

- [tex]\( np = 20 \times 0.59 = 11.8 \[/tex] ) (greater than 10, so this condition is met)

-[tex]$ n(1-p) = 20 \times 0.41 = 8.2 $[/tex] (less than 10, so this condition is not met)

Since one of the conditions is not met, we use a simulation to assess the probability.

Evaluate the simulation results:

- From the simulation, we know that 446 out of 2000 samples had a sample proportion of 0.70 or higher.

- This gives us an empirical probability: [tex]$ \frac{446}{2000} = 0.223 $[/tex] or 22.3%.

Interpret the result:

The empirical probability of obtaining a sample proportion of 0.70 or higher, assuming the true proportion is 0.59, is 22.3%.

This probability is not very low, suggesting that while a sample proportion of 0.70 is not extremely rare, it is not highly unlikely either.