Respuesta :
Answer:
The 90% confidence interval for the difference between means is (-161.18, 205.18).
Step-by-step explanation:
Sample mean and standard deviation for Region I:
[tex]M=\dfrac{1}{12}\sum_{i=1}^{12}(438+1013+1127+737+...+1075+500+340)\\\\\\ M=\dfrac{8424}{12}=702[/tex]
[tex]s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{12}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}\cdot [(438-(702))^2+(1013-(702))^2+...+(500-(702))^2+(340-(702))^2]}\\\\\\[/tex]
[tex]s=\sqrt{\dfrac{1}{11}\cdot [(69696)+(96721)+...+(131044)]}\\\\\\s=\sqrt{\dfrac{1174834}{11}}=\sqrt{106803.1}\\\\\\s=326.8[/tex]
Sample mean and standard deviation for Region II:
[tex]M=\dfrac{1}{15}\sum_{i=1}^{15}(778+464+563+...+479+710+389+826)\\\\\\ M=\dfrac{10204}{15}=680[/tex]
[tex]s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{15}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{14}\cdot [(778-(680))^2+(464-(680))^2+...+(389-(680))^2+(826-(680))^2]}\\\\\\[/tex]
[tex]s=\sqrt{\dfrac{1}{14}\cdot [(9551.804)+(46771.271)+...+(84836.27)+(21238.2)]}\\\\\\ s=\sqrt{\dfrac{545975.7}{14}}=\sqrt{38998}\\\\\\s=197.5[/tex]
Now, we have to calculate a 90% confidence level for the difference of means.
The degrees of freedom are:
[tex]df=n1+n2-2=12+15-2=25[/tex]
The critical value for 25 degrees of freedom and a confidence level of 90% is t=1.708
The difference between sample means is Md=22.
[tex]M_d=M_1-M_2=702-680=22[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{326.8^2}{12}+\dfrac{197.5^2}{15}}\\\\\\s_{M_d}=\sqrt{8899.853+2600.417}=\sqrt{11500.27}=107.24[/tex]
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=1.708 \cdot 107.24=183.18[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 22-183.18=-161.18\\\\UL=M_d+t \cdot s_{M_d} = 22+183.18=205.18[/tex]
The 90% confidence interval for the difference between means is (-161.18, 205.18).
