Question

Need to use Rstudio for this question

A government grants is funding a study to calculate how long it takes for the average consumer to establish an Internet connection. A random sample of 20 Internet users’ connection time is collected. The connections time in seconds are

0.03, 0.48, 0.49, 0.52, 0.66, 0.67, 0.70, 0.76,0.82, 1.2, 1.22, 1.39, 1.62, 1.85, 1.97, 2.25, 2.84, 3.11, 3.48, 4.02

Use chi-square test to see if it is reasonable to assume that internet connection time follows an exponential distribution with a mean of 1.5 seconds.

\begin{tcolorbox} Hint:

• You need to generate 20 samples from exponential distribution using rexp function

• The dexp function in R uses rate in its argument, not mean. You need to revisit Week2 lecture note to find the conversion between rate and mean of exponential distribution.

• You will need to bin both expected and observed data before the test. You can use hist() function in R to bin the data. For example

#generate 100 random samples from normal distribution with mean 0

# and standard deviation of 1

set.seed(1234)

d<-rnorm(100, 0, 1)

z1<-hist(d)

z1$breaks

# shows the breaks of each bin. For example, the first bin contains values between -2.5 and -2.0

z1$counts

# frequency counts

The function hist may use different bin size for different data, therefore you will need to specify the breaks argument in the hist function.

Answer 1

Statistics and Probability

Need to use R-studio for this question

A government grants is funding a study to calculate how long it takes for the average consumer to establish an Internet connection. A random sample of 20 Internet users’ connection time is collected. The connections time in seconds are 0.03, 0.48, 0.49, 0.52, 0.66, 0.67, 0.70, 0.76, 0.82, 1.2, 1.22, 1.39, 1.62, 1.85, 1.97, 2.25, 2.84, 3.11, 3.48, 4.02.

Use chi-square test to see if it is reasonable to assume that internet connection time follows an exponential distribution with a mean of 1.5 seconds.

Answer:

Using chi-square test to see if it is reasonable to assume that internet connection time follows an exponential distribution with a mean of 1.5 seconds.

We have chi square p-value= 0.7167531. Hence it is concluded that internet connection time follows an exponential distribution with a mean of 1.5 seconds.

R code is below:

time<-c(0.03, 0.48, 0.49, 0.52, 0.66, 0.67, 0.70, 0.76,0.82, 1.2, 1.22, 1.39, 1.62, 1.85, 1.97, 2.25, 2.84, 3.11, 3.48, 4.02) #################### Chi square test ################### x.exp.cut<-cut(time,breaks=c(0, 1, 2, 3, 4.5)) ##binning data, identify breaks by hist(time$breaks) observed<-table(x.exp.cut) ## binned data table observed ## computing expected frequencies mean=1.5 E1<-(pexp(1,rate=1/mean)-pexp(0,rate=1/mean))*length(time) E2<-(pexp(2,rate=1/mean)-pexp(1,rate=1/mean))*length(time) E3<-(pexp(3,rate=1/mean)-pexp(2,rate=1/mean))*length(time) E4<-(pexp(4.5,rate=1/mean)-pexp(3,rate=1/mean))*length(time) expected<-c(E1,E2,E3,E4) ## expected frequencies vector chi<-sum((observed-expected)^2/expected) df<-4-1 1-pchisq(chi,df) ## p-value #[1] 0.7167531 # p-value chi-square