A) Does it seem reasonable to assume that the sample ?x = (x1,...,xn), the obser

Question

A) Does it seem reasonable to assume that the sample ?x = (x1,...,xn),

the observed values of X1, . . . , Xn, were drawn from:

(a) a normal distribution? Why or why not?

(b) a symmetric distribution? Why or why not?

B) Assume that X1, . . . , Xn are normally distributed and let ? = EXi = ?.

(a) Test the null hypothesis derived above using Student’s 1-sample t-test. What is the significance probability? If we adopt a signif- icance level of ? = 0.05, should we reject the null hypothesis?

(b) Construct a (2-sided) confidence interval for ? with a confidence coefficient of approximately 0.90.

Explanation / Answer

A)

a)

Here we do shapiro wilk's normaility test for both the categories Cross and Self

We do this test in R.

The R code is given below:

x=c(23.5,12,21,22,19.1,21.5,22.1,20.4,18.3,21.6,23.3,21,22.1,23,12)
y=c(17.4,20.4,20,20,18.4,18.6,18.6,15.3,16.5,18,12.8,15.5,18)
shapiro.test(x)
shapiro.test(y)

Now, we get the following results in R

Shapiro-Wilk normality test

data: x
W = 0.7529, p-value = 0.0009707

data: y
W = 0.9294, p-value = 0.3352

Hence from the p value we can conclude that Self fertilised are normally distributed but cross fertilised are not normally distributed

b)To check symmetricity we installed a package called "lawstat"

R code:
library(lawstat)
symmetry.test(x)
symmetry.test(y)

We get the following results

data: x
Test statistic = -2.3718, p-value = 0.058
alternative hypothesis: the distribution is asymmetric.

y
Test statistic = -0.8359, p-value = 0.43
alternative hypothesis: the distribution is asymmetric.

hence at 5% level of significance we conclude that both the datas are symmetric as p value>0.05

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.