MATH SOLVE

2 months ago

Q:
# The article "Occurrence and Distribution of Ammonium in Iowa Groundwater" (K. Schilling, Water Environment Research, 2002:177β186) describes measurements of ammonium concentrations (in mg/L) at a large number of wells in the state of Iowa. These included 349 alluvial wells and 143 quaternary wells. Of the alluvial wells, 182 had concentrations above 0.1, and 112 of the quaternary wells had concentrations above 0.1. Find a 95% confidence interval for the difference between the proportions of the two types of wells with concentrations above 0.1.

Accepted Solution

A:

Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).Step-by-step explanation:Since we have given that Number of alluvial wells = 349Number of quaternary wells = 143Number of alluvial wells that had concentrations above 0.1 = 182Number of quaternary wells that had concentrations above 0.1 = 112Average of alluvial wells = 0.27Standard deviation = 0.4Average of quaternary wells = 1.62 Standard deviation =1.70So, 95% confidence interval gives alpha = 5% level of significance.[tex]\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96[/tex]So, 95% confidence interval becomes,[tex](1.62-0.27)\pm 1.96\sqrt{\dfrac{0.4^2}{349}+\dfrac{1.7^2}{143}}\\\\=1.35\pm 1.96\times 0.020\\\\=(1.35-0.040,1.35+0.040)\\\\=(1.31,1.39)[/tex]Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).