91Ó°ÊÓ

It is given that $c=90\%=0.90$

$\hat{p}=0.36,n=1416$

For confidence interval$1-\mathrm{Î\pm }=0.90$,$z_{a/2}=1.96$using probability table.

The conditions are:

Random: US adults are selected randomly, it is satisfied.

Normal: Success are $1416(0.36)=509.76$and failure are $1416(1-0.36)=906.24$which are larger than ten.

Independent: $1416$adults is surely less than $10\%$of total population.

All conditions are satisfied.

Now,

Margin of Error: $E=z_{\mathrm{Î\pm }/2}×\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=1.96×\sqrt{\frac{0.36(1-0.36)}{1416}}=0.0250$

Confidence Interval: $\hat{p}-E=0.6-0.0250=0.3350$

and $\hat{p}+E=0.6+0.0250=0.3850$

hence, interval is$(0.3350,0.3850)$

The interval does not have $0.5$. All values are below $0.5$. It shows that population proportion is less than $0.5$. Hence, there is convincing evidence that less than half of adults can name all branches of government.