Question

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.

Answer 1

Answer:

$z=\frac{0.14 -0.11}{\sqrt{\frac{0.11(1-0.11)}{1630}}}=3.871$  

Critical approach

$z_{crit}= 1.64$

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:

$p_v =P(z>3.871)=0.000054$  

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

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

$p_o=0.11$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ 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:$p\leq 0.11$  

Alternative hypothesis:$p >0.11$  

The statistic for this case is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.14 -0.11}{\sqrt{\frac{0.11(1-0.11)}{1630}}}=3.871$  

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:

$z_{crit}= 1.64$

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:

$p_v =P(z>3.871)=0.000054$  

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.