HWK 8: Statistical Inferencea. You have the following data of wind speed (in mph) in Ithaca, NY at 3 pmeach day:10, 7, 3, 9, 7, 10, 12, 13, 1, 12.Evaluate whether this data is a better match for the normal or lognormaldistribution using the Chi Square test, the Maximum Likelyhood test, and theKalmogorov-Smirnoff test. Use the winning distribution to predict theprobability of getting wind speeds over 15 mph. It is recommended you useR for these calculations, but show what you are doing in yourhwksubmission.

Solution: Given data x=c(10,7,3,9,7,9,12,13,1,12) By using R-software install.packages("EnvStats")library(EnvStats)x=c(10,7,3,9,7,9,12,13,1,12) ############# chi-square test (for normal) ######################### R code is as follows: gofTest(x,distribution="norm",est.arg.list=list(method="mle"),test="chisq") ##################And the output Results of Goodness-of-Fit Test------------------------------- Test Method: Chi-square GOF Hypothesized Distribution: Normal Estimated Parameter(s): mean = 8.300000sd = 3.716181 Estimation Method: mle/mme Data: x Sample Size: 10 Test Statistic: Chi-square = 3.2 Test Statistic Parameter: df = 3 P-value: 0.361805 Alternative Hypothesis: True cdf does not equal theNormal Distribution.Warning message:In chisqGofTest(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), n.classes = NULL, :Expected counts < 5. Chi-squared approximation may notbe appropriate. ############ KS- test (for normal) ######################### R code is as follows: gofTest(x,distribution="norm",est.arg.list=list(method="mle"),test="ks") ##################And the output is:################# Results of Goodness-of-Fit Test------------------------------- Test Method: Kolmogorov-Smirnov GOF Hypothesized Distribution: Normal Estimated Parameter(s): mean = 8.300000sd = 3.716181 Estimation Method: mle/mme Data: x Sample Size: 10 Test Statistic: ks = 0.1747049 Test Statistic Parameter: n = 10 P-value: 0.9203193 Alternative Hypothesis: True cdf does not equal theNormal Distribution.Warning messages:1: In ksGofTest(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), alternative = "two.sided", :The standard Kolmogorov-Smirnov test is very conservative (Type I error smaller than assumed; high Type II error) for testing departures from the Normal distribution when you have to estimate the distribution parameters. 2: In ks.test(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), y = "pnorm", :ties should not be present for the Kolmogorov-Smirnov test ########## chi-square test (for lognormal) ######################### R code is as follows: gofTest(x,distribution="lnorm",est.arg.list=list(method="mle"),test="chisq") ##################And the output is:################# Results of Goodness-of-Fit Test------------------------------- Test Method: Chi-square GOF Hypothesized Distribution: Lognormal Estimated Parameter(s): meanlog = 1.9222229sdlog = 0.7560852 Estimation Method: mle/mme Data: x Sample Size: 10 Test Statistic: Chi-square = 11.6 Test Statistic Parameter: df = 3 P-value: 0.008886889 Alternative Hypothesis: True cdf does not equal theLognormal Distribution.Warning message:In chisqGofTest(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), n.classes = NULL, :Expected counts < 5. Chi-squared approximation may notbe appropriate. ######### KS- test (for lognormal) ######################### R code is as follows: gofTest(x,distribution="lnorm",est.arg.list=list(method="mle"),test="ks") ##################And the output is:################# Results of Goodness-of-Fit Test------------------------------- Test Method: Kolmogorov-Smirnov GOF Hypothesized Distribution: Lognormal Estimated Parameter(s): meanlog = 1.9222229sdlog = 0.7560852 Estimation Method: mle/mme Data: x Sample Size: 10 Test Statistic: ks = 0.3124963 Test Statistic Parameter: n = 10 P-value: 0.2828932 Alternative Hypothesis: True cdf does not equal theLognormal Distribution.Warning messages:1: In ksGofTest(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), alternative = "two.sided", :The standard Kolmogorov-Smirnov test is very conservative (Type I error smaller than assumed; high Type II error) for testing departures from the Lognormal distribution when you have to estimate the distribution parameters. 2: In ks.test(x = c(10, 7, 3, 9, 7, 9, 12, 13, 1, 12), y = "plnorm", :ties should not be present for the Kolmogorov-Smirnov test From p-values of output we conclude that the data is good fit to normal distribtionwith parameter mean= 8.3 and S.D=3.716181 Now to probability of wind speed over 15 mph is given by 1-pnorm(15,8.3,3.716181) Output is : 0.03569