Question

The built-in data set treering provides Annual tree-ring widths in normalized units for years from -6000 to 1979. Assume that the n measurements x=(

x₁, x₂,...,x_n

) are a random sample from a population true mean μ and true unknown variance

σ²

. Using R we can define the vector x by the assignment x<-as.vector(treering).
a) Calculate, n, the number of elements in x.  

b)Calculate the sample standard deviation s, of x.  

c) Estimate true mean μ, using this data by calculating the sample mean.  

d) Calculate an unbiased point estimate of the population variance,

σ²

of tree-ring widths.  

e) Assuming normality of tree ring widths, calculate the maximum likelihood estimate of μ?  
f) Calculate the 60th percentile of x using R.  

g) Calculate a

1

798

trimmed mean for x using R.  

h) Since the sample size is >30 we can create a confidence interval for μ using a normal critical value. If we want the confidence interval to be at the 96% level and we use a normal critical value, then what critical value should we use?

i) Calculate a 96% confidence interval(using a normal critical value) for μ.
(

,

)

j) How long is the 96% confidence interval just created in part i?  

Answer 1

Rcode for first 4 bits

data(treering)
head(treering)
x<-as.vector(treering)
print(x)
no_of_elements <- length(x)
no_of_elements
standard_deviation <- sd(x)
standard_deviation
sample_mean <- mean(x)
sample_mean
population_variance <- var(x)
population_variance

Ouptut:

> no_of_elements
[1] 7980
> standard_deviation <- sd(x)
> standard_deviation
[1] 0.3003575
> sample_mean <- mean(x)
> sample_mean
[1] 0.9968362
> population_variance <- var(x)
> population_variance
[1] 0.09021466

Solution-A:

number of elements=7980

Solution-B:

sample standard deviation=s=0.3003575

Solution-c:

sample mean=0.9968362

Solutiond:

unbiased point estimate of the population variance,

σ^{2=s^2=0.09021466}