91Ó°ÊÓ

We need to find shape of sampling distribution.

As given ,
$$\begin{gathered} n_1=100 \\ {n}_2=70 \\ {p}_1=80\%=0.80 \\ {p}_2=60\%=0.60 \end{gathered}$$
It is safe to assume that the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$ is approximately normal if,
$n_1{p}_1â‰\yen 10,n_1\left(1-p_1\right)â‰\yen 10,n_2{p}_2â‰\yen 10,n_2\left(1-p_2\right)â‰\yen 10.$
Thus, we have,
$$\begin{gathered} n_1{p}_1=100(0.80)=80 \\ {n}_1(1-p_1)=100(1-0.80)=100(0.20)=20 \\ {n}_2{p}_2=70(0.60)=42 \\ {n}_2(1-p_2)=70(1-0.60)=70(0.40)=28 \end{gathered}$$
All the conditions are met, which implies that the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$is approximately normal.

We need to find mean of sampling distribution.

It is given in the question that:
$$\begin{gathered} n_1=100 \\ {n}_2=70 \\ {p}_1=80\%=0.80 \\ {p}_2=60\%=0.60 \end{gathered}$$
The mean of the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$ is the difference between the proportions:
${\mathrm{μ}}_{\widehat{p_1}-\widehat{p_2}}=p_1-p_2=0.80-0.60=0.20$
The mean is $0.20$

We need to find the standard deviation of the sampling distribution.

It is given in the question that:
$$\begin{gathered} n_1=100 \\ {n}_2=70 \\ {p}_1=80\%=0.80 \\ {p}_2=60\%=0.60 \end{gathered}$$

The mean of the sampling distribution of is the difference between the proportions:

${\mathrm{μ}}_{\widehat{p_1}-\widehat{p_2}}=p_1-p_2=0.80-0.60=0.20$

The standard deviation of ${\hat{p}}_1-{\hat{p}}_2$ is given by:
$$\begin{gathered} {\mathrm{σ}}_{\widehat{p_1}-\widehat{p_2}}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}} \\ =\sqrt{\frac{0.80(1-0.80)}{100}+\frac{0.60(1-0.60)}{70}} \\ =0.09586 \end{gathered}$$
Thus, we conclude that the difference between Nathan's state and Kyle's state in the sample proportion of cars made by the American manufacturers varies on average by $0.09586$from the mean difference $0.20$.