Answer:
Step-by-step explanation:
The null and the alternative hypothesis is:
[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
Using the data of t-value and degree of freedom;
The P-value = -0.224
Decision rule: Do not 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 ∝.
Therefore, there is enough evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United States.
