Question

To compare water quality from wells in two different areas, water samples were taken from 50 wells in the first area, tested, and 31 were found to meet standards. From the second area, 42 out of 60 wells were found to meet standards. Use the p-value method to test the claim that both areas have the same proportion of wells that meet standards at the 95% confidence level. Based on this result, does there appear to be a significant difference in water quality between the two regions?

Answer 1

To Test :-

H0 :- P1 = P2
H1 :- P1 ≠ P2

p̂1 = 31 / 50 = 0.62
p̂2 = 42 / 60 = 0.7
Test Statistic :-
Z = ( p̂1 - p̂2 ) / √( p̂ * q̂ * (1/n1 + 1/n2) ))
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 31 + 42 ) / ( 50 + 60 )
p̂ = 0.6636
q̂ = 1 - p̂ = 0.3364
Z = ( 0.62 - 0.7) / √( 0.6636 * 0.3364 * (1/50 + 1/60) )
Z = -0.8843

Decision based on P value
P value = 2 * P ( Z < -0.8843 ) = 0.3766
Reject null hypothesis if P value < α = 0.05
Since P value = 0.3766 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There does not appears to be a significant difference in water quality between the two regions.