A collection of data samples is independent if they come from unrelated populations and the samples do not affect each other.Using the Kruskal-Wallis Test, we can decide whether the population distributions are identical without assuming them tofollow the normal distribution. In the built-in data set named airquality, the daily air quality measurements in New York,May to September 1973, are recorded. The ozone density is presented in the data frame column Ozone. Without assuming thedata to have normal distribution, test at .05 significance level if the monthly ozone density in New York has identical datadistributions from May to September 1973. An R code and its output is given below.R code and its output> head (airquality)Ozone Solar.R Wind Temp Month Day190 7.4 675 15 241236118 8.072.....> kruskal.test (Ozone - Month, data = airquality)Kruskal-Wallis rank sum testdata: Ozone by MonthKruskal-Wallis chi-squared = 29.267, df = 4, p-value6.901e-06Which of the following is the correct interpretation of the result of data analysis using R?I. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 arenonidentical populations.II. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 areidentical populations.III. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 arenot significantly different.a. I onlyb. П onlyс. Ш onlyd. Both II and IIIO DO BO A

The study is about to test the monthly ozone density in New York has identical data distribution…

A collection of data samples is independent if they come from unrelated populations and the samples do not affect each other. Using the Kruskal-Wallis Test, we can decide whether the population distributions are identical without assuming them to follow the normal distribution. In the built-in data set named airquality, the daily air quality measurements in New York, May to September 1973, are recorded. The ozone density is presented in the data frame column Ozone. Without assuming the data to have normal distribution, test at .05 significance level if the monthly ozone density in New York has identical data distributions from May to September 1973. An R code and its output is given below. R code and its output > head (airquality) Ozone Solar.R Wind Temp Month Day 41 36 5 1 5 2 190 7.4 67 2 118 8.0 72 ..... > kruskal.test (Ozone - Month, data = airquality) Kruskal-Wallis rank sum test data: Ozone by Month Kruskal-Wallis chi-squared - 29.267, df = 4, p-value - 6.901e-06 Which of the following is the correct interpretation of the result of data analysis using R? I. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 are nonidentical populations. II. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 are identical populations. III. At .05 significance level, we conclude that the monthly ozone density in New York from May to September 1973 are not significantly different. a. I only b. П оnly c. III only d. Both II and III

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 19PFA