g According to the U.S. Census Bureau, 11% of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30, 2011). Suppose that in a recent sample of 1630 children, 228 were found to be living with at least one grandparent. At a 5% significance level, can you conclude that the proportion of all children in the United States who currently live with at least one grandparent is higher than 11%? Use both the p-value and the critical-value approaches.

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true proportion is significantly higher than 0.11 or 11%

P value

We are conducting a right tailed test so then the p value is given by:

[tex]p_v =P(z>3.871)=0.000054[/tex]

And we see that is a very low value compared to the significance level of 0.05 so then we have enough evidence to conclude that the true proportion is significantly higher than 0.11.

Step-by-step explanation:

Information provided

n=1630 represent the random sample selected

X=228 represent the children were found to be living with at least one grandparent

[tex]\hat p=\frac{228}{1630}=0.140[/tex] estimated proportion of children were found to be living with at least one grandparent

[tex]p_o=0.11[/tex] is the value to verify

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic

[tex]p_v[/tex] represent the p value

System of hypothesis

We want to verify if the % of children who live with at least one grandparent is higher than 11%, so then the system of hypothesis is .:

Null hypothesis:[tex]p\leq 0.11[/tex]

Alternative hypothesis:[tex]p >0.11[/tex]

The statistic for this case is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

Replacing the info given we got:

[tex]z=\frac{0.14 -0.11}{\sqrt{\frac{0.11(1-0.11)}{1630}}}=3.871[/tex]

Critical approach

we need to find a critical value in the normal standard distribution who accumulate 0.05 of the area in the left and for this case this value is:

[tex]z_{crit}= 1.64[/tex]

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true proportion is significantly higher than 0.11 or 11%

P value

We are conducting a right tailed test so then the p value is given by:

[tex]p_v =P(z>3.871)=0.000054[/tex]

And we see that is a very low value compared to the significance level of 0.05 so then we have enough evidence to conclude that the true proportion is significantly higher than 0.11.