Question

The mayor of a city wants to see if pollution levels are reduced by closing the streets to the car traffic. This is measured by the rate of pollution every 60 minutes (8am 22pm: total of 15 mea- surements) in a day when traffic is open, and in a day of closure to traffic. Here the values of air pollution: With traffic: 214, 159, 169, 202, 103, 119, 200, 109, 132, 142, 194, 104, 219, 119, 234 Without traffic: 159, 135, 141, 101, 102, 168, 62, 167, 174, 159, 66, 118, 181, 171, 112 (a) Is this paired data or independent samples? (b) Construct the proper test to see if the pollution levels are reduced by closing the streets to the car traffic. Give the test statistic and the p-value. (c) Since the p-value depends on sample size, some people may wish to calculate an effect size which can be done dividing the absolute (positive) Standardize test statistic z by the square root of the number of pairs. R does not give the standardize test statistic in its output but it can be calculated using the p-value from the output and the following formula: abs (qnorm (pvalue))/2. According to Cohen's classification of effect sizes which is 0.1 (small effect), 0.3 (moderate effect) and 0.5 and above (large effect). What is the effect size from this test? (d) Compute a 95% confidence interval for the difference between the true pollution levels.

Answer 1

Que.a

This is paired data. Because we measure pollution in the same city with different situations (treatment) namely with and without car traffic.

Que.b

We use paired sample t test to see if the pollution levels are reduced by closing the streets to the car traffic.

Hypothesis:

I used R software to solve this question.

R codes and output:

> traffic=c( 214, 159, 169, 202, 103, 119, 200, 109, 132, 142, 194, 104, 219, 119, 234)
> wo_traffic=c(159, 135, 141, 101, 102, 168, 62, 167, 174, 159, 66, 118, 181, 171, 112)
> t.test(traffic,wo_traffic,alternative='greater',paired=T)

Paired t-test

data: traffic and wo_traffic
t = 1.5128, df = 14, p-value = 0.07629
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-4.413266 Inf
sample estimates:
mean of the differences
26.86667

Test statistic = t = 1.5128

p-value = 0.07629

Since p-value is greater than 0.05, we accept null hypothesis.

Que.c

> z=abs (qnorm (0.07629))/2
> effect=z/sqrt(length(traffic))
> effect
[1] 0.1846739

Effect size is 0.18, which is small effect.

Que.d

> t.test(traffic,wo_traffic,paired=T)

Paired t-test

data: traffic and wo_traffic
t = 1.5128, df = 14, p-value = 0.1526
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-11.22361 64.95694
sample estimates:
mean of the differences
26.86667

95 percent confidence interval for the difference between the true pollution levels is ( -11.22361 , 64.95694 )