Question 1

The following questions are based on confidence intervals.

(a)  [4 Marks] Let, (2.00, 3.26, 2.84, 4.77, 3.23, 3.64, 1.84, 4.43, 1.35, 2.28)  N (µ, σ2 ) with µ ∈ R and σ 2  is unknown. Calculate a 90% confidence interval for µ .

Solution:

rm(list = ls())

set.seed(100)

x <- rnorm(10, mean = 3, sd = 2)

round(x, 2)

## 90% CI for the mu ##

gam <- 0.9

xbar <- mean(x); xbar

s2 <- var(x)

SE <- sqrt(s2/length(x))

ll <- xbar - qt((1+gam)/2, df = length(x) - 1) * SE

ul <- xbar + qt((1+gam)/2, df = length(x) - 1) * SE

paste0("90% CI is:", " (", round(ll, 2), ", ", round(ul, 2), ")")

[1] "90% CI is: (2.31, 3.61)"

Thus the 90% CI is = (2.31, 3.61)

(b)  [5 Marks] Using the same data set provided in (a), calculate the left sided 95% confidence interval for σ 2 .

Solution:

The shortest confidence interval here will be the left side confidence interval.  Thus the lower limit is 0. We just need to calculate the upper limit,

## 95% shortest CI for the mu ##

def <- length(x) - 1

gam <- 0.95

chis <- qchisq(1 - gam, df = def)

ll <- 0

ul <- def*s2 /chis

paste0("95% CI is:", " (", round(ll, 2), ", ", round(ul, 2), ")")

[1] "95% CI is: (0, 8.82)"

[1] ”95% CI is: (0, 3.41)”

Question 2

(a)  [5 Marks] A tow sided 95% confidence interval for µ in the location Normal model N(µ, σ0(2)), calculated from a an observed sample is (50, 60). ( Note: sample size is not given to you)

Calculate the lower limit of the 80% confidence interval for µ, using the data from the same sample.

Solution

 =  =  = 55

z(1+0.80)/2  = 1.282

ME = U L  = 60 50  = 5

σ0              ME  = 2.55102

Lower limit of the 80% confidence interval

N(=)ot(x¯)do(.80)t(2)ed(=) n(5)5 fo(8)r(×)th(2.)ti(=)on(51) .73469     ■

(b)  [8 Marks] Using R, I generated a sample of 64 observations from a Poisson(λ) distribution and calculated an approximate two sided 95% confidence interval for λ using the approximation based on the central limit theorem.  The lower limit of this confidence interval is L = 2.64. Calculate the upper limit of this confidence interval.

Solution

L =  − z √   =⇒  − √ − L = 0

z   ± √  +4L          1 .96 ± √  +4 ×2 .64

■

Question 3

Suppose that x = (1, 1, 0, 0, 1) is an observed sample from Bernoulli(θ) distribution and the prior

distribution of θ has p.d.f.:

π(θ) = {0(5)θ4 ,

(a)  [5 Marks] Calculate the posterior expectation of θ 2 .

Solution: Note that the prior distribution of θ is beta(5,1) and thus the posterior distribution is beta(5+3 = 8, 1 + 2 = 3)

π(θ | s) = θ 7 (1 − θ)2

Thus, the posterior expectation of θ 2 ,

E(θ2  | s) =   θ 9 (1 − θ)2 dθ =   = 0.5454545

Note for TA: Please check for any potential error in the solution.

(b)  [5 Marks] Calculate the posterior expectation of  .

Solution: Thus, the posterior expectation of 1/θ ,

E(θ−1  | s) =   θ 6 (1 − θ)2 dθ =   = 1.428571

Question 4

(a)  [5 Marks] I generated a sample from a location Normal model with σ0(2)  = 16. The 95% confidence interval for the mean (µ) calculated using this sample I generated is (58.7, 61.9). This interval has all the information necessary to test the null hypothesis H0  : µ = 59 against the alternative H1  : µ > 59. Calculate the value of the observed z-statistic and the critical z-value for the z-test of the null hypothesis H0  : µ = 59 against the alternative H1  : µ > 59. Assume α = 0.05 The 95% confidence interval for the mean (µ) calculated using this sample I generated is (58.7, 61.9) =⇒ SE =  =2(61)  = 0.8163265

 =  = 60.3

z = 0    = 068(0)1(.)6(3)2(5)6(9)5  = 1 .59

The critical z-value at α = 0 .05 is Z0 .95  = 1 .645

Since, the calculated z 1.59 < 1.645, thus, we cannot reject the H0        ■

(b)  [3 Marks] Calculate the number of observations in this study.

Solution: Let the number of observations = n. We know that

SE = σ0 / √n =2 ⇒ n = (σ0 /0.8163265)2  = 24.01 ≈ 25