Question

Question 1. Consider the following datapoints r1, .. . , Tg: 1.12,2.09,1.49, 1.63,2.76. 1.41,1.64, 3.41 1 marlk 1 marlk 1 mark (d) Suppose the r, above are observations of an independent random sample X1,... , Xn from a N(μ, σ*) distribution. Using as a pivot, construct a 95% confidence (a) Compute c. (b) Compute the observed sample variance s2 (c) Compute the median of the xis. interval for μ. One of the following facts is helpful ·P(-1.96 < T < 1.96) = 0.95 if T ~ N(0, 1) ·P(-2.36 < < 2.36) 0.95 if T ~ t ·P(218 < T < 17.53) 0.95 iIT ~ χ . ·P(-2.26 < T < 2.26)-0.95 if T ~ t9 7. marks (e) Suppose that, in the situation of part (d), we wanted a confidence interval for σ2 instead of μ. What would be an appropriate pivot, and what is its distribution? 2 marks

Answer 1

### By using R command

> x=c(1.12,2.09,1.49,1.63,2.76,1.41,1.64,3.41)
> x
[1] 1.12 2.09 1.49 1.63 2.76 1.41 1.64 3.41
> xbar=mean(x) ### Sample mean
> xbar
[1] 1.94375
> Ssq=var(x) ##Sample variance
> Ssq
[1] 0.5996554
> S=sqrt(Ssq) ##Sample standard deviation
> S
[1] 0.7743742
> t=2.36
> UCl=(xbar-t*(S/sqrt(7)) ## upper confidence limit

> UCl
[1] 1.253011
> LCl=(xbar+t*(S/sqrt(7))) ## lower confidence limit
> LCl
[1] 2.634489
> CI=c(UCl,LCl) ## Confidence interval
> CI
[1] 1.253011 2.634489
>

d) If we wanted a confidence interval for the population variance we used following Pivot:

$T=\frac{(n-1)S^2}{\sigma ^2}\sim \chi _{n-1}$