PLEASE SHOW WORK..TEST USED IN EXCEL ETC... The service times below are believed

Question

PLEASE SHOW WORK..TEST USED IN EXCEL ETC...

The service times below are believed to come from a process that can be approximated by a continuous triangular distribution with parameters (5, 10, 15). Service times: 2.3, 9.0, 7.5, 11.9, 12.1, 12.0, 8.2, 7.0, 7.1, 14.3, 0.7, 9.0, 11.4, 6.7, 12.4, 9.6, 5.3, 10.8, 8.7, 8.0. (a) Draw the CDF of the both the triangular distribution and the sampled data on the same graph (b) Formulate a hypothesis test and calculate the test statistic for the KS-test. (c) For a sample size of 20, and level of significance of .01, the critical value for the hypothesis test is a = .294. What conclusions can be drawn?

Explanation / Answer

I use R software to solve this

(a)

rm(list=ls(all=TRUE))
library(triangle)
x=c(2.3, 9,7.5,11.9,12.1,8.2,7,7.1,14.3,0.7,9,11.4,6.7,12.4,9.6,5.3,10.8,8.7,8)
x1=rltriangle(length(x), 5,15,10)
c1=pltriangle(sort(x),5,15,10)
c2=pltriangle(sort(x1),5,15,10)
plot(sort(x),c1, type="l")
lines(sort(x1),c2, type="l",lty=2)
legend("bottomright",c("TRI CDF","Emprical CDF"),lty=1:2)

(b)

The test hypothesis is

H0: The data follow a specified distribution

H1: The datado not follow a specified distribution

R-command

ks.test(x, "pltriangle")

One-sample Kolmogorov-Smirnov test

data: x
D = 0.35831, p-value = 0.01521
alternative hypothesis: two-sided

(c) The p-value is greater than the level of significance, we concluded that do not reject the null hypothesis and the data follow a same distribution.

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