An engineer on the staff of a water purification plant measu

An engineer on the staff of a water purification plant measures the chlorine content in 200 different samples daily.Over the period of years, she has established the population standard deviation to be 1.4 milligrams of chlorine per liter.The latest samples averaged 4.6 milligrams of chlorine per liter.

Find the standard error of the mean.

Establish the symmetric interval around 5.2, the population mean, which will include the sample mean with a probability of 68.3 percent.

Solution

a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=1.4
Sample Size(n)=200
Standard Error = ( 1.4/ Sqrt ( 200) )
= 0.099
b)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=5.2
Standard deviation( sd )=1.4
Sample Size(n)=200
Confidence Interval = [ 5.2 ± Z a/2 ( 1.4/ Sqrt ( 200) ) ]
= [ 5.2 - 1 * (0.099) , 5.2 + 1 * (0.099) ]
= [ 5.101,5.299 ]