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 = 349
Number of quaternary wells = 143
Number of alluvial wells that had concentrations above 0.1 = 182
Number of quaternary wells that had concentrations above 0.1 = 112
Average of alluvial wells = 0.27
Standard deviation = 0.4
Average of quaternary wells = 1.62
Standard deviation =1.70
So, 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).