A large study involved 540 household For each household the

A large study involved 540 household. For each house-hold, the concentration of arsenic in the water (in mu g/l) was measured. The data are given in the file ArsenicConcentrations.txt. The table below gives the mean and standard deviation for the arsenic concentration measurements and their natural logarithm. Figure 16.1 gives the histogram of the concentrations of arsenic (on the left), and the histogram of the log-concentrations of arsenic (on the right). Give point estimates for the mean and the geometric mean of the concentration of arsenic. Construct a 95% confidence interval for the mean concentration o arsenic. Compute a 95% confidence interval for the mean log-concentration arsenic. By exponentiating the limits of this interval, obtain a 95% conference interval for the geometric mean of the concentration of arsenic. Comparing the confidence intervals from (b) and (c), which one do your think it describes better the center of the distribution of concentrations arsenic? Why?

Solution

Solution:

best point estimate =sample mean=0.8441

Geometric mean=exponential of mean of logx=e^0.8441 =2.3259

Solutionb:

Given sample large samples

Follws SRS

and normal distribution.

Z for 95%1.96

95% confidence level for population mean=

=0.8441±1.96(2.1014/sqrt(524)

=0.8441±0.1799

0.8441-0.1799<<0.8441+0.1799

0.6642<<1.024

lower imit=0.6642

upper limit=1.024

we are 95% confident that the true population mean concentrtion of arsenic lies in between 0.6642 and 1.024

SolutionC:

As given in problem

log concentration mean=-1.3193

log concentration sd=1.5200

95% confidence interval for geometic mean of arsenic

-1..3193±1.96(1.5200/sqrt(524)

=-1.3193±0.1301

-1.3193-0.1301< mu<-1.3193+0.1301

-1.4494 <mu<-1.1892

lower limit=-1.4494

upper limit=-1.1892