Question

\The dataset below contains the concentration of NO2 measured hourly measured at twenty randomly selected monitoring stations across Australia in 2013. According to the National Air Quality Standards1 , it is a concern if the NO2 concentration exceeds 0.12 ppm. An environmentalist wished to know if there is a concern or not. (a) Based on the information provided, write down the null and alternative hypotheses that the environmentalist would like to test. [1 mark] (b) Write down the name of the test you are going to use. [1 mark] (c) Write down the decision rule. [1 mark] (d) Paste the output from R Commander. (If you would like to do it by hand, calculate the test statistic and show the steps.) [1 mark] (e) Do you reject the null hypothesis or not? Explain your answer. [1 mark] (f) Draw a conclusion which answers the original problem. [1 mark] (g) State and check all the assumptions for the hypothesis test you have used. [3 marks] Note: Parts (a) to (f) are actually the six steps of hypothesis testing. You may use it as a guideline for some of the remaining questions.

NO2_2013
0.108
0.122
0.165
0.194
0.116
0.084
0.095
0.141
0.15
0.146
0.126
0.126
0.132
0.068
0.101
0.061
0.304
0.116
0.052
0.165

Answer 1

a) The null and alternative hypotheses are Ho 0.12 H1: μ > 0.12 (b) one sample t test (c) Decision Rule: If p-value is less than 0.05, reject the null hypothesis (d) The output from R Commander using Data set name as NO2 is > with (NO2, (.cest (NO2_2013, alternative-'greater',mu-0.12, + Conf.level-.95))) One Sample t-test data: NO2 2013 t = 0.69969, df = 19, p-value = 0.2463 alternative hypothesis: true mean is greater than 0.12 95 percent confidence interval: 0.1073471 Inf sample estimates: an of x 0.1286 (e) Do not reject the null hypothesis (f) The concentration of NO2 does not exceeds 0.12 ppm (g) The assumptions of one sample t test are 1. Dependent variable must be continuous and normally distributed 2. The observations are independent of one other The NO2 measurements are continuous and the observations are independent. We need to check for the normality assumption.

The output in R commander to check the normality is: > with (NO2, Hist (N02-2013, scale-frequency", col="darkgray") ) breaks="Sturges", 0 寸 0.05 0.10 0.15 0.20 0.25 0.30 0.35 N02-2013 It looks like there exist an outier hence the data is not normally distributed therefore, the assumption of normality is not satisfied.