Answer:
Step-by-step explanation:
The null and the alternative hypothesis can be computed as:
[tex]H_o: \mu_{c} \ge \mu_{us}[/tex]
[tex]H_a: \mu_{c} < \mu_{us}[/tex]
The t- student test statistics can be computed as:
[tex]t = \dfrac{ x_c - x_{us} }{\sqrt { \dfrac{\sigma_c^2}{n_c} + \dfrac{\sigma_{us} ^2}{n_{us}} } }[/tex]
[tex]t = \dfrac{ 19 - 21 }{\sqrt { \dfrac{7^2}{100} + \dfrac{8^2}{130} } }[/tex]
t = -2.017
degree of freedom = (n₁-1)+(n₂-1)
= (100-1) + (130-1)
= 228
At t = -2.017 and 228 df
P-value = -0.224
Decision rule: Not to reject the null hypothesis if the p-value is greater than ∝(0.01)
Conclusion: We reject the null hypothesis since the p-value is less than ∝.
Thus, there is sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United states.
