Question

Independent random samples from two regions in the same area gave the following chemical measurements (ppm). Assume the population distributions of the chemical are mound-shaped and symmetric for these two regions.
Region I: ; 438 1013 1127 737 491 840 306 402 1155 1075 500 340
Region II: ; 778 464 563 610 827 894 476 394 824 387 816 767 479 710 389 826

Required:
Find a 90% confidence interval.

Answer 1

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:

$M=\dfrac{1}{12}\sum_{i=1}^{12}(438+1013+1127+737+...+1075+500+340)\\\\\\ M=\dfrac{8424}{12}=702$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{12}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}\cdot [(438-(702))^2+(1013-(702))^2+...+(500-(702))^2+(340-(702))^2]}$

$s=\sqrt{\dfrac{1}{11}\cdot [(69696)+(96721)+...+(131044)]}\\\\\\s=\sqrt{\dfrac{1174834}{11}}=\sqrt{106803.1}\\\\\\s=326.8$

Sample mean and standard deviation for Region II:

$M=\dfrac{1}{15}\sum_{i=1}^{15}(778+464+563+...+479+710+389+826)\\\\\\ M=\dfrac{10204}{15}=680$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{15}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{14}\cdot [(778-(680))^2+(464-(680))^2+...+(389-(680))^2+(826-(680))^2]}$

$s=\sqrt{\dfrac{1}{14}\cdot [(9551.804)+(46771.271)+...+(84836.27)+(21238.2)]}\\\\\\ s=\sqrt{\dfrac{545975.7}{14}}=\sqrt{38998}\\\\\\s=197.5$

Now, we have to calculate a 90% confidence level for the difference of means.

The degrees of freedom are:

$df=n1+n2-2=12+15-2=25$

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.

$M_d=M_1-M_2=702-680=22$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{326.8^2}{12}+\dfrac{197.5^2}{15}}\\\\\\s_{M_d}=\sqrt{8899.853+2600.417}=\sqrt{11500.27}=107.24$

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.708 \cdot 107.24=183.18$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 22-183.18=-161.18\\\\UL=M_d+t \cdot s_{M_d} = 22+183.18=205.18$

The 90% confidence interval for the difference between means is (-161.18, 205.18).