Question

#2. You've sampled acorns from m^2 quadrates in Maine (the data are below). Based on these data, answer the following questions.

no.acorns<-c(3,26,11,24,22,16,15,26,20,22,17,16,18,20,18,15,17,16,12,25,22,20,27,19,13,17,17,20,25,26,13,21,12,27,19,15,14,20,17,15,15,23,27,17,17,14,18,23,18,23)

#A What is an appropriate distribution to use to model this data?

#B What is/are the parameter estimate(s) for this distribution?

Answer 1

Solution:

Part A

The histogram for given data by using R is given as below:

no.acorns = c(3,26,11,24,22,16,15,26,20,22,17,16,18,20,18,15,17,16,12,25,22,20,27,19,13,17,17,20,25,26,13,21,12,27,19,15,14,20,17,15,15,23,27,17,17,14,18,23,18,23)

> hist(no.acorns)

From above histogram, it is observed that data follows an approximately normal distribution.

Now, we have to conduct Kolmogorov Smirnov test for checking the normality of the given data.

This test by using R is given as below:

R commands and outputs are given as below:

> ks.test(no.acorns, "pnorm", mean=mean(no.acorns), sd=sd(no.acorns))

        One-sample Kolmogorov-Smirnov test

data: no.acorns

D = 0.093198, p-value = 0.7779

alternative hypothesis: two-sided

Here, we get p-value > alpha 0.05, so we do not reject the null hypothesis that given data follows normal distribution.

So, we conclude that given data follow an approximately normal distribution.

Part B

Now, we have to find the estimates for parameters for this distribution which are given as below:
> mean(no.acorns)

[1] 18.66

> sd(no.acorns)

[1] 4.934717

Mean = 18.66

SD = 4.9347