91Ó°ÊÓ

The given values are,

$$\begin{gathered} x_1=215, \\ {n}_1=300, \\ {x}_2=186, \\ {n}_2=250, \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

The null hypothesis

$H_0:p_1=p_2$

The alternative hypothesis

$H_a:p_1>p_2$

Using MINITAB output.

MINITAB output: Test and $\mathrm{CI}$two proportions

From the MINITAB, the test statistic is $-0.72$and the $p-$value is $0.473.$

$p$- value is more than the level of significance.

$p$- value $(=0.473)<a(=0.05)$

Using the rejection rule, it can be concluded that there is evidence to reject null hypothesis $H_0$at $alpha=0.005.$

Therefore, at $5\%$significance level, the data do not provide sufficient evidence to conclude that there is a difference between the labor-force participation rates of U.S. and Canadian women.

The given values are,

$$\begin{gathered} x_1=215, \\ {n}_1=300, \\ {x}_2=186, \\ {n}_2=250, \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

Using MINITAB output.

Using the MINITAB output.

MINITAB output: Test and CI for two proportions.

From the MINITAB output, $95\%$confidence interval is $(-0.101675,0.0470087)$. Therefore, there is $95\%$confidence interval for the difference between the proportion is $(-0.101675,0.0470087)$