Question

Exercise 5, Below are suminary statistics for n = 155 lead concentration measurements (in ppm) collected from the flooded banks of the Meuse river. A plot of the data (x and y coordinates) is also shown below (note that the data set can be accessed from the sp package). Calculate a 90% confidence interval for the mean lead concentration in this region. Are any statistical assumptions for the confidence interval you constructed not satisfied? library(sp) data("meuse") summary (meuse$lead) ## Min. 1st Qu. Median 37.0 72.5 123.0 153.4 207.0 654.O Mean 3rd Qu Max sd (meuse$lead) ## [1] 111.3201 coordinates (meuse) <- xtV bubble (meuse, "lead", maxsize-1.25, main-"Lead Concentration (ppm) ") Lead Concentration (ppm) es. So. 37 72.5 123 207 ● 654

Answer 1

we have,

$\bar{x}=153.4$

$s=111.3201$

the 90% confidence interval for population mean in this case is given by,

$[\bar{x}-\frac{s}{\sqrt{n}}t_{\alpha/2,n},\bar{x}+\frac{s}{\sqrt{n}}t_{\alpha/2,n}]$

where   $t_{\alpha/2,n}$ is the upper $\alpha/2th$ point of the t- distribution with n observations.

in our case, $\alpha=0.10\implies \alpha/2=0.05$

from the t-table we get,

$t_{0.05,155}=1.345$

Therefore the required CI is,

$[153.4-\frac{111.3201}{\sqrt{155}}*1.645,153.4+\frac{111.3201}{\sqrt{155}}*1.645]$

$=[153.4-8.9414*1.645,153.4+8.9414*1.645]$

$=[153.4-14.71,153.4+14.71]$

$=[138.69,168.11]$

The Statistical assumptions that the sample comes from a normal distribution and the standard deviation of this normal distribution is unknown are not satisfied for the constructed confidence interval.