Data Analysis

Basic Problem

• There is a population whose properties we are interested in and wish to quantify statistically: mean, standard deviation, distribution, etc.

• The Question – Given a sample, what was the random system that generated its statistics?

Central Limit Theorem

• If one takes random samples of size n from a population of mean and standard deviation then as n gets large, approaches the normal distribution with mean and standard deviation

• is generally unknown and often replaced by the sample standard deviation s resulting in , which is termed the Standard Error of the sample.

n

X

ns

Example

Normal distribution

Critical Values for Confidence Levels

1- .80 .90 .95 .99 0.20 0.10 0.05 0.01/2 0.10 .05 .025 0.005z/2 1.28 1.64 1.96 2.58

Confidence Interval for Mean(large sample size)

)(2/ nszX

)(2/ XSEzX

OR

Student’s t-distribution

Critical Values for Confidence Levelst-distribution

1- .80 .90 .95 .99 0.20 0.10 0.05 0.01/2 0.10 .05 .025 0.005d.f. = 1 3.09 6.31 12.71 63.66d.f. = 10 1.37 1.81 2.23 4.14d.f. = 30 1.31 1.70 2.04 2.75d.f. = 1.28 1.65 1.96 2.58

Confidence Interval for Mean(small sample size, t-distribution)

)(2/ nstX

)(2/ XSEtX

OR

Comparing Population Means

)( 212/2121 XXSEtXX

2

22

1

21

21 )(n

s

n

sXXSE

2121

11)(

nnsXXSE pool

Pooled Variance

Unequal Variance

Hypothesis Testing (t-test)

• Null Hypothesis – differences in two samples occurred purely by chance

• t statistic = (estimated difference)/SE• Test returns a “p” value that represents the

likelihood that two samples were derived from populations with the same distributions

• Samples may be either independent or paired

Tails

• One tailed test – hypothesis is that one sample is: less than, greater than, taller than,

• Two tailed test – hypothesis is that one sample is different (either higher or lower) than the other

Paired Test

• Samples are not independent• Much more robust test to determine

differences since all other variables are controlled

• Analysis is performed on the differences of the paired values

• Equivalent to Confidence interval for the mean

Paired Samples – New Site

TSS Concentrations vs. Time

BMP Performance Comparison

• Commonly expressed as a % reduction in concentration or load– Highly dependent on influent concentration– Potentially ignores reduction in volume (load)

• May lead to very large differences in pollutant reduction estimates

• Preferable to compare discharge concentrations

Effect of TSS Influent Concentration

-100%

-80%

-60%

-40%

-20%

0%

20%

40%

60%

80%

100%

0 100 200 300 400 500 600

Influent Concentration

% R

emov

al

Swale

Strip

Detention

Sand Filter

Wet Pond

Sand Filter - TSS

y = 0.0046x + 7.4242

R2 = 0.0037

0

50

100

150

200

250

300

350

400

0 100 200 300 400 500

TSS Influent (mg/L)

TSS

Eff

luen

t (m

g/L)

Comparison of Effluent Quality

Exercise

• Calculate average concentrations for each constituent for the two watersheds

• Determine whether any concentrations are significantly different, report p value for null hypothesis

• Calculate average effluent concentrations for the two BMPs and determine whether they are different from the influent concentrations – p values

• Compare effluent concentrations for the two BMPs and determine whether one BMP is better than the other for a particular constituent.