Question

Concentrations of lead in groundwater greater than 120 ppm is considered unhealthy. 36 samples of a particular source of water are to be collected and analyzed, and the null hypothesis H0 u<= 120 ppm will be tested against the alternative hypothesis Ha: u > 120ppm . Let X be the lead concentration in a sample. The standard deviation for the concentration of lead in groundwater is known to be 16ppm .          
a.   A decision is made to reject H0 if x >= 125ppm . Find the significance level of the test.

b.   Find the power of the test described in part (a) if the true mean concentration is 128 ppm.

c.   How many samples should be analyzed to have power 0.99 when the true mean concentration is 128 ppm if the test is conducted at the 3% level of significance?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
mean μ=	120
standard deviation σ=	16

std error=σx̅=σ/√n=16/√36=

2.6667

P(sig level)=P(X>125|µ=120) =P(X>125)=P(Z>(125-120)/2.667)=P(Z>1.88)=1-P(Z<1.88)=1-0.9699=0.0301

b)

Power=P(X>125|µ=128) =P(X>125)=P(Z>(125-128)/2.667)=P(Z>-1.13)=1-P(Z<-1.13)=1-0.1292=0.8708

c)

Hypothesized mean μo=		120
true mean μa=		128
std deviation σ=		16.00
0.03 level critical Z=		1.88
0.01 level critical Zβ=		2.33
n=(Zα/2+Zβ)2σ2/(μo-μa)2=		71