Definition

Random Vector

• Let \(\mathbf X_{p\times 1}=(X_1, \cdots, X_p)^T\) be a \(p\)-dimensional random vector. Its mean vector and covariance are defined as follows:

\[\begin{align*} \boldsymbol \mu & =E[\mathbf X]=\begin{pmatrix} E[X_1]\\ \vdots \\ E[X_p] \end{pmatrix}\\ \boldsymbol \Sigma_{p\times p} &= Cov(\mathbf X) = E[(\mathbf X -\boldsymbol \mu)(\mathbf X - \boldsymbol \mu)^T] \end{align*}\]

A Linear Combination of a Random Vector

• Consider a linear combination of \(\mathbf X\):

\[Y=\mathbf a^T \mathbf X,\] where \(\mathbf a\) is a \(p\times 1\) vector of constants, i.e., \(a=(a_1, \cdots, a_p)^T\).

• \(Y=\mathbf a^T \mathbf X =\sum_{i}a_i X_i = a_1 X_1 + \cdots + a_p X_p\).

• \(Y\) is a random variable, which is a linear combination of the random vector \(\mathbf X\). It is a univariate random variable.

Mean

Mean of \(Y=a^TX\)

• The mean of \(Y\) can be expressed as: \[ \begin{aligned} E(Y) &= E(\mathbf{a}^T\mathbf{X}) \\ &= \mathbf{a}^T E(\mathbf{X}) \\ &= \mathbf{a}^T \boldsymbol{\mu} \end{aligned} \]

• Intuitively, the mean of \(Y\) is a weighted average of the components of \(\mathbf{X}\), with weights given by the corresponding components of \(\mathbf{a}\).

Variance

Variance of \(Y\)

• The variance of \(Y\) can be expressed as: \[ \begin{aligned} \text{Var}(Y) &= \text{Var}(\mathbf{a}^T\mathbf{X}) \\ &= E[(\mathbf{a}^T\mathbf{X}-\mathbf{a}^T\boldsymbol\mu)^2]\\ &=E[(\mathbf{a}^T\mathbf{X}-\mathbf{a}^T\boldsymbol\mu)(\mathbf{a}^T\mathbf{X}-\mathbf{a}^T\boldsymbol\mu)^T]\\ &=E[\mathbf{a}^T(\mathbf{X}-\boldsymbol\mu)(\mathbf{X}-\boldsymbol\mu)^T\mathbf a]\\ &= \mathbf{a}^T \boldsymbol{\Sigma} \mathbf{a} \end{aligned} \]

Variance of \(Y\) and Quadratic Forms

• The variance of \(Y\) depends on the covariance structure of \(\mathbf{X}\), as well as the weights given by \(\mathbf{a}\).

• We call forms like \(\mathbf a^T \boldsymbol \Sigma \mathbf a\) as quadratic forms.

• Note, we can also write the variance of \(Y\) as:

Quadratic Forms

• A quadratic form is a polynomial of degree 2 in the components of a vector, and it can be expressed as: \[Q(\mathbf{X}) = \mathbf{X}^T \mathbf{A} \mathbf{X}\] where \(\mathbf{A}\) is a symmetric matrix.
• We will discuss distributions of certain quadratics forms later.

Summary: Mean and Var of \(Y=\mathbf a^T \mathbf X\)

• If \(\mathbf X\sim (\boldsymbol \mu, \boldsymbol \Sigma)\), then \(Y=\mathbf a^T\mathbf X\sim (\mathbf a^T\boldsymbol \mu, \mathbf a^T\boldsymbol \Sigma \mathbf a)\).

• \(\mathbf a^T \boldsymbol \mu\) is the inner product between \(\mathbf a\) and \(\boldsymbol \mu\), \[\sum_{i=1}^p a_i \mu_i.\]

• \(\mathbf a^T \boldsymbol \Sigma \mathbf a\) is a quadratic form: \[\mathbf a^T \boldsymbol \Sigma \mathbf a=\sum_{i=1}^p\sum_{j=1}^p a_i a_j \sigma_{ij},\] where \(\sigma_{ij}\) is the \((i,j)\)th element of \(\boldsymbol \Sigma\).

Estimation of Mean and Variance of \(Y\)

• Recall that if \(X_1, \cdots, X_n\overset{iid}\sim (\boldsymbol \mu, \boldsymbol \Sigma)\), then

  • \(\bar{X}\sim (\boldsymbol \mu, \frac{1}{n}\boldsymbol \Sigma)\)
  • \(E[\mathbf S]=\boldsymbol \Sigma\) where \[\mathbf S=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})(X_i-\bar{X})^T.\]

• We can estimate its mean \(\mathbf a^T \boldsymbol \mu\) by \(\mathbf a^T \bar{\mathbf X}\), which is a linear combination of the sample mean vector.

• We can estimate its variance, which is \(\mathbf a^T \boldsymbol \Sigma \mathbf a\), by \(\mathbf a^T \mathbf S \mathbf a\), which is a quadratic form of the sample covariance matrix.

Example 1

• \(\mathbf X = (X_1, X_2)^T\), \(a=(1/2, 1/2)^T\).

Then \[Y=a^TX=\frac{1}{2}(X_1 + X_2)\]

• \(E(Y)=\frac{1}{2}(\mu_1 + \mu_2)\).

• \(Var(Y) = \frac{1}{4} (\sigma_1^2 + \sigma_2^2 + 2\sigma_{12})= \sum_{i=1}^2 \sum_{j=1}^2 a_i a_j \sigma_{ij}\). Note that \(\sigma_{12}=\sigma_{21}\).

Example 2

• Assume we have a random sample from a distribution with mean \(\mu\) and variance \(\sigma^2\), i.e., \(X_1, \cdots, X_n\overset{iid}\sim (\mu, \sigma^2)\).

• We often stack the random variables vertically: \[\mathbf X_{n\times 1}=\begin{pmatrix} X_1 \\ \vdots \\ X_n\end{pmatrix}.\]

• An equivalent expression, \(\mathbf X=(X_1, \cdots, X_n)^T\).

Example 2

• The random vector \(\mathbf X\) formed by a random sample from a distribution with mean \(\mu\) and variance \(\sigma^2\) has the following mean vector and covariance matrix:

\[\begin{align*} E[\mathbf X] &=\mu \mathbf 1_n =\begin{pmatrix}\mu \\ \vdots \\ \mu\end{pmatrix}\\ Cov(\mathbf X) &=\sigma^2 \mathbf I_n = \begin{bmatrix} \sigma^2 & 0 & \cdots & 0 \\ 0 & \sigma^2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma^2 \end{bmatrix} \end{align*}\]

Example 2

• We can express the sample mean as a linear combination of the random vector \(\mathbf X\): \[\bar X= \frac{1}{n}X_1 + \cdots + \frac{1}{n}X_n=\frac{1}{n}\mathbf 1^T \mathbf X,\] where \(\mathbf 1=(1, \cdots, 1)^T\) is a \(n\times 1\) vector.
• By the linear combination results, we have \[\begin{align*} E[\bar X]&=\frac{1}{n} \mathbf 1^T E[\mathbf X]=\frac{1}{n} \mathbf 1^T \mu \mathbf 1=\mu\\ Var[\bar X]&=\frac{1}{n^2} \mathbf 1^T Cov(\mathbf X) \mathbf 1=\frac{1}{n^2} \mathbf 1^T \sigma^2 \mathbf I_n \mathbf 1 = \frac{1}{n}\sigma^2. \end{align*}\]

Example 2

• \(\mathbf 1\) vs \(\mathbf 1^T\):
  • \(\mathbf 1\) is a \(n\times 1\) vector, which is a column vector.
  • \(\mathbf 1^T\) is a \(1\times n\) vector, which is a row vector.
• \(\mathbf 1^T \mathbf 1\) vs \(\mathbf 1 \mathbf 1^T\):
  • inner product: \(\mathbf 1^T \mathbf 1\) is a \(1\times 1\) scalar, which is equal to \(n\).
  • outer product: \(\mathbf 1 \mathbf 1^T\) is a \(n\times n\) matrix, which is a matrix of all ones. We often denote it by \(\mathbf J\), i.e., \(\mathbf J=\mathbf 1 \mathbf 1^T\)

Example 3

Linear Combinations of Iris Setosa Features

• Recall that for the iris setosa, \(\mathbf X\) is \(50\times 4\).
• Consider a linear combination of the features \(Y= \mathbf Xb\), where

\[ b=\begin{pmatrix} 1/4 \\ 1/4 \\ 1/4 \\ 1/4 \end{pmatrix} \]

• \(Yb\) is a \(50\times 1\) vector, with the \(i\)th row be the average of the four features of the \(i\)th iris setosa flower. To see this

Linear Combinations of Iris Setosa Features

\[\begin{aligned} Y&=\mathbf Xb= \begin{pmatrix} X_1^T \\ \vdots \\ X_n^T \end{pmatrix} b = \begin{pmatrix} X_1^Tb \\ \vdots \\ X_n^Tb \end{pmatrix} =\begin{pmatrix} b^TX_1 \\ \vdots \\ b^TX_n \end{pmatrix}\\ & =\begin{pmatrix} \frac{x_{11} +x_{12} + x_{13} + x_{14}}{4} \\ \vdots \\ \frac{x_{n1} +x_{n2} + x_{n3} + x_{n4}}{4} \end{pmatrix} \end{aligned} \]

Linear Combinations of Iris Setosa Features: sample mean

setosa=as.matrix(iris[iris$Species=="setosa", 1:4])
sample.meanvec=matrix(colMeans(setosa), 4, 1)
sample.cov=cov(setosa)
b=matrix(1/4, 4, 1)
Y=setosa%*%b
#sample mean of Y: the following two results are the same
mean(Y)

[1] 2.5355

t(b)%*%sample.meanvec

       [,1]
[1,] 2.5355

Linear Combinations of Iris Setosa Features: sample variance

#sample variance of Y: the following two results are the same
var(Y)

           [,1]
[1,] 0.03844617

t(b)%*%cov(setosa)%*%b

           [,1]
[1,] 0.03844617

The Mean of a Univariate Distribution

• A random sample \(X_1, \cdots, X_n\) from a univariate distribution with mean \(\mu\) and variance \(\sigma^2\).
• We are interested in making inference about the population mean \(\mu\).
• The sample mean \(\bar X\) is an unbiased estimator of \(\mu\), i.e., \(E[\bar X]=\mu\). We often use \(\hat\mu=\bar X\) to estimate \(\mu\).
• How to quantify the uncertainty of \(\hat\mu\)? Recall that \(Var(\bar X)=\frac{\sigma^2}{n}\)..

Variance, Sample Variance, and Standard Error

• \(\sigma^2\) is unknown. But the sample variance \(s^2\) is an unbiased estimator of \(\sigma^2\), i.e., \(E[s^2]=\sigma^2\). We often use \(\hat\sigma^2=s^2\) to estimate \(\sigma^2\).

• The sample variance \(s^2\) is defined as \(s^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)^2\)

• The standard error of \(\bar X\) is defined as \(se(\bar X)=\sqrt{Var(\bar X)}=\frac{\sigma}{\sqrt{n}}\).

• We can estimate it by \(se(\bar X)=\frac{s}{\sqrt{n}}\).

Confidence Intervals for \(\mu\)

• A ``large-sample’’ (approximate) confidence interval for \(\mu\) is given by \[\bar X \pm z_{\alpha/2} se(\bar X)\] where \(z_{\alpha/2}\) is the upper \(\alpha/2\) quantile of the standard normal distribution.
• A ``small-sample’’ (approximate) confidence interval for \(\mu\) is given by \[\bar X \pm t_{\alpha/2} se(\bar X)\] where \(t_{\alpha/2}\) is the upper \(\alpha/2\) quantile of the t-distribution with \(n-1\) degrees of freedom.

Linear Combinations of Means

• In many situations, the parameter of interest is a function of the means.

• For example, we may be interested in the mean of a linear combination of the means, i.e., \(\mathbf a^T \boldsymbol \mu = \sum_{i=1}^p a_i \mu_i\), where \(\mathbf a=(a_1, \cdots, a_p)^T\) is a \(p\times 1\) vector.

• In the following simulated study, we will show how to construct a large-sample confidence interval for \(\mathbf a^T \boldsymbol \mu\).

A Random Sample From a Multivariate Distribution

• Consider a random sample \(\mathbf X_1, \cdots, \mathbf X_n\) from a multivariate distribution with mean vector \(\boldsymbol \mu_{p\times 1}\) and covariance matrix \(\boldsymbol \Sigma_{p\times p}\).
• We often stack the random vectors to form an \(n\times p\) matrix: \[\mathbf X_{n\times p}=\begin{pmatrix}\mathbf X_1^T \\ \vdots \\ \mathbf X_n^T\end{pmatrix}\]

Sample Mean Vector

• Sample mean vector is \[\bar{\mathbf X}=\frac{1}{n}\sum_{i=1}^n \mathbf X_i=(\frac{1}{n}\mathbf 1_n^T \mathbf X)^T\]

It is a random vector with

• mean vector \(E[\bar{\mathbf X}]=\boldsymbol \mu\), i.e., the sample mean vector is unbaised for the population mean vector. \(\bar{\mathbf X}\) can be used to estimate \(\boldsymbol \mu\).

• covariance matrix \(Cov(\bar{\mathbf X}) = \frac{1}{n} \boldsymbol \Sigma\)

Sample Covariance Matrix

• The sample covariance matrix is \[\mathbf S = \frac{1}{n-1}\sum_{i=1}^n(\mathbf X_i-\bar{\mathbf X})(\mathbf X_i-\bar{\mathbf X})^T\]

• It is unbiased for \(\boldsymbol \Sigma\), i.e., \(E[\mathbf S]=\boldsymbol \Sigma\).

• We showed that \[\mathbf S= \frac{1}{n-1} \mathbf X^T \mathbf C \mathbf X \] where \(\mathbf C_{n\times n}=\mathbf I-\frac{1}{n}\mathbf J=\mathbf I-\frac{1}{n}\mathbf 1 \mathbf 1^T\)

• This expression is helpful when we derive the distribution of \(\mathbf S\).

Inference of \(\mathbf a^T \boldsymbol \mu\)

• Two basic questions:

  • How to estimate it?

  • What is the standard error of the estimator?

• We have shown that \(\bar{\mathbf X}\sim (\boldsymbol \mu, \frac{1}{n}\boldsymbol \Sigma)\), which implies that \[\mathbf a^T \bar{\mathbf X} \sim (\mathbf a^T \boldsymbol \mu, \mathbf a^T \frac{1}{n}\boldsymbol \Sigma \mathbf a)\]

• An unbiased estimator of \(\mathbf a^T \boldsymbol \mu\) is \(\mathbf a^T \bar{\mathbf X}\), which is a linear combination of the sample mean vector.

Standard Error of \(\mathbf a^T \bar{\mathbf X}\)

• The standard error of \(\mathbf a^T \bar{\mathbf X}\) is defined as \(se(\mathbf a^T \bar{\mathbf X})=\sqrt{Var(\mathbf a^T \bar{\mathbf X})}=\sqrt{\mathbf a^T \frac{1}{n}\boldsymbol \Sigma \mathbf a}\).

• We can estimate it by \(se(\mathbf a^T \bar{\mathbf X})=\sqrt{\mathbf a^T \frac{1}{n}\mathbf S \mathbf a}\).

• Confidence intervals can be constructed by \(\mathbf a^T \bar{\mathbf X} \pm z_{\alpha/2} se(\mathbf a^T \bar{\mathbf X})\) for large-sample C.I. and \(\mathbf a^T \bar{\mathbf X} \pm t_{\alpha/2} se(\mathbf a^T \bar{\mathbf X})\) for small-sample C.I.

• We will explain exact and asymptotic distributions of \(\mathbf a^T \bar{\mathbf X}\) later.

Description and Parameters

• This is a simulated data set

• For adults, the recommended range of daily protein intake is between 0.8 g/kg and 1.8 g/kg of body weight

• 60 observations

• 4 sources of proteins

  • meat

  • dairy

  • vegetables / nuts / tofu

  • other

Choose Mean Vector and Covariance Matrix

• The multivariate distribution has
  • mean vector \[\boldsymbol \mu=\begin{pmatrix}24, 16, 8, 8\end{pmatrix}^T\]
  • covariane matrix \[\boldsymbol \Sigma=4* \begin{pmatrix} 1.3 & 0.3 & 0.3 & 0.3\\ 0.3 & 1.3 & 0.3 & 0.3\\ 0.3 & 0.3 & 1.3 & 0.3\\ 0.3 & 0.3 & 0.3 & 1.3 \end{pmatrix}\]

Define Mean Vector and Covariance Matrix in R

#the library "MASS" is required
library(MASS)
my.cov=4*(diag(4) + 0.3* rep(1,4)%o%rep(1,4))
eigen(my.cov)#to check whether the cov matrix is p.d.

eigen() decomposition
$values
[1] 8.8 4.0 4.0 4.0

$vectors
     [,1]       [,2]       [,3]       [,4]
[1,] -0.5  0.8660254  0.0000000  0.0000000
[2,] -0.5 -0.2886751 -0.5773503 -0.5773503
[3,] -0.5 -0.2886751 -0.2113249  0.7886751
[4,] -0.5 -0.2886751  0.7886751 -0.2113249

my.mean=8*c(3,2,1,1)
n=60

Generate a random sample from the multivariate distribution

Simulate A Random Sample

set.seed(1)
x=mvrnorm(n, mu=my.mean, Sigma=my.cov)
dim(x)

[1] 60  4

protein=as.matrix(data.frame(meat=x[,1],dairy=x[,2], 
                             veg=x[,3], other=x[,4]))

The simulated data

protein

          meat    dairy       veg     other
 [1,] 29.08891 17.54865  5.814221  7.264953
 [2,] 23.65965 13.06336  8.734581  9.452868
 [3,] 26.43410 16.83504  9.278807  8.409798
 [4,] 21.68232 15.51922  3.379171  5.954558
 [5,] 22.22387 15.45446  8.804571  7.562144
 [6,] 25.54395 16.46835  8.556332 10.299174
 [7,] 20.15075 14.71290 10.660378  7.584075
 [8,] 25.44330 14.98680  4.866275  6.323171
 [9,] 23.41142 16.34138  6.667006  6.164109
[10,] 28.21604 16.64242  5.874860  7.078538
[11,] 22.58127 13.61817  5.178349  5.652878
[12,] 22.19211 16.04745  8.714666  6.732854
[13,] 25.97926 16.80008  7.189986  9.716474
[14,] 25.66703 20.61869 13.775770  9.078236
[15,] 20.16010 16.09623  5.020107  8.049388
[16,] 24.57145 18.88263  6.894722  5.917792
[17,] 23.25621 14.96338 10.680367  7.196099
[18,] 22.60198 16.38243  5.220357  6.195492
[19,] 22.91070 15.01628  7.664447  5.536308
[20,] 22.09802 15.96519  6.882419  7.530778
[21,] 21.65197 16.70303  8.420131  3.772608
[22,] 22.60577 11.60921  9.084612  8.060032
[23,] 25.92991 15.29974 10.415539  3.912418
[24,] 24.31179 20.74766 11.504271 11.239025
[25,] 24.10939 18.66505  3.604761  5.943392
[26,] 24.65994 13.87394 12.148044  5.651083
[27,] 26.07243 12.43699  7.787358 10.627558
[28,] 25.65462 17.71194 11.203610 10.155746
[29,] 25.35010 17.99635  9.018100  6.472298
[30,] 23.84272 16.53127  8.723904  4.422476
[31,] 21.04508 13.96795  5.848427  7.077553
[32,] 26.24455 14.99075  8.126523  7.248014
[33,] 25.43487 15.58447  6.258134  6.422490
[34,] 25.29261 18.33085  5.907034  6.788726
[35,] 28.79099 17.70326 11.313192  6.362595
[36,] 25.58286 15.77977 11.144694  5.954819
[37,] 22.37370 16.52364  8.412213 11.029752
[38,] 23.09505 18.58350  6.704652  7.968698
[39,] 20.24731 15.93243  8.673596  4.620260
[40,] 22.04807 13.03450  6.280768 10.108768
[41,] 23.16952 18.11268  5.688636 10.005273
[42,] 24.44874 16.62458  8.455707  7.974149
[43,] 21.38847 14.99773  6.050585  9.428152
[44,] 23.44805 14.45326  6.170624  8.625406
[45,] 23.88782 19.45005  7.836918  8.911576
[46,] 28.11042 11.39956  9.940819 10.746740
[47,] 24.70061 15.72228  6.630922  6.783132
[48,] 24.43655 17.83334  4.404197  4.766235
[49,] 24.83206 15.88387  8.316900  7.633714
[50,] 25.60672 13.17837  6.049305  5.938031
[51,] 22.30839 14.24987  5.246096 11.833706
[52,] 24.10819 20.38798  4.572755 10.562201
[53,] 25.97482 14.87898  5.463695  7.658656
[54,] 24.54808 17.48112 10.983295  9.687974
[55,] 21.51529 14.44885  6.041175  5.492619
[56,] 20.38223 13.44285  5.149195  5.276100
[57,] 23.99043 15.17236  9.281141  9.734778
[58,] 25.06526 16.68357  6.961285 13.484693
[59,] 24.01093 12.41687  8.165424  8.026655
[60,] 23.89317 14.91410  7.783749 10.210253

Sample Mean and Sample Covariance

xbar=matrix(colMeans(protein), 4, 1)
t(xbar)

         [,1]     [,2]    [,3]     [,4]
[1,] 24.03403 15.92836 7.66049 7.738634

S=cov(protein)
S

           meat     dairy       veg     other
meat  4.2956426 0.8150757 1.1294478 0.5532420
dairy 0.8150757 4.4052993 0.3497889 0.2337300
veg   1.1294478 0.3497889 5.1705794 0.5897121
other 0.5532420 0.2337300 0.5897121 4.5287293

Estimation

• An unbiased estimator of \(\boldsymbol\mu\) is the sample mean vector, i.e., \(\hat{\boldsymbol \mu}=\bar{\mathbf X}\).
• An unbiased estimator of \(\boldsymbol \Sigma\) is the sample covariance matrix \(\mathbf S\), i.e., \(\hat{\boldsymbol \Sigma}=\mathbf S\)
• We have shown that \(Var(\bar{\mathbf X})=\frac{1}{n}\boldsymbol \Sigma\), where \(n=60\).
• We can estimate it by \[\hat{Var}(\bar{\mathbf X})=\frac{1}{60}\mathbf S\]

Linear Functions/Combinations: Three Questions

• Suppose we only have a random sample and we would like to make inference of the following:

• Q1: Construct a large-sample (approximate) C.I. for protein from meat. In other words, the parameter of interest is \(\mu_1\).

• Q2: Construct a large-sample C.I. for the total protein intake

• Q3: Construct a large-sample C.I. for the difference of protein intake between from meat and from vegetable

Linear Functions/Combinations: Question 2

• Q2: Construct a large-sample C.I. for the total protein intake.

• The parameter of interest is \(\mu_1+\mu_2+\mu_3+\mu_4=\mathbf a^T \boldsymbol \mu\), where \(\mathbf a=(1,1,1,1)^T\).

• Estimate: \(\mathbf a^T \bar{\mathbf X}\)

• Standard error: \[\sqrt{\mathbf a^T\frac{\mathbf S}{n} \mathbf a}\]

Linear Functions/Combinations: Question 3

• Q3: Construct a large-sample C.I. for the difference of protein intake between from meat and from vegetable
• The parameter of interest is \(\mu_1 - \mu_3=\mathbf a^T \boldsymbol \mu\), where \(\mathbf a=(1, 0, -1, 0)^T\).
• Estimate: \(\mathbf a^T \bar{\mathbf X}\)
• Standard error: \(\sqrt{\mathbf a^T\frac{\mathbf S}{n} \mathbf a}\)

R Code

Question 1

• R code to compute using the above two ways

sqrt(S[1,1]/60) # Method 1

[1] 0.2675706

# Method 2
a=matrix(c(1,0,0,0),4,1)
sqrt(t(a)%*%S%*%a/60)

          [,1]
[1,] 0.2675706

• Both methods give \(s.e.(\bar X_{(1)})=0.27\)

• An approximate 95% C.I. for \(\mu_1\) is \(24.0 \pm 1.96*0.27\)

Question 2

a=matrix(1,4,1)
t(a)%*% xbar #estimate 

         [,1]
[1,] 55.36152

sqrt(t(a)%*%S%*%a/60) #standard error

          [,1]
[1,] 0.6550095

#a large-sample 95% C.I. 
c(t(a)%*% xbar- 1.96*sqrt(t(a)%*%S%*%a/60), 
  t(a)%*% xbar+ 1.96*sqrt(t(a)%*%S%*%a/60))

[1] 54.07770 56.64534

Question 3

a=matrix(c(1,0,-1,0),4,1)
t(a)%*% xbar #estimate 

         [,1]
[1,] 16.37354

sqrt(t(a)%*%S%*%a/60) #standard error

          [,1]
[1,] 0.3465864

#a large-sample 95% C.I. 
c(t(a)%*% xbar- 1.96*sqrt(t(a)%*%S%*%a/60), 
  t(a)%*% xbar+ 1.96*sqrt(t(a)%*%S%*%a/60))

[1] 15.69423 17.05285

knitr::knit_exit()