Setup

Throughout let $X_1,\dots ,X_n\stackrel{\mathrm{ind}}{\sim }\mathcal{N}(\mu ,{\sigma }^2)$.

In this lecture we review how to:

1. Estimate $\mu$ and ${\sigma }^2$.
2. Build confidence intervals for $\mu$ and ${\sigma }^2$.
3. Test hypotheses concerning $\mu$ and ${\sigma }^2$.
4. Choose the sample size.

We call $X_1,\dots ,X_n$ a random sample.

Golden ratio example:

A class of $n=27$ students measured $B/A$ on themselves:

gr <- c(1.66, 1.61, 1.62, 1.69, 1.58, 1.43, 1.66, 
        1.69, 1.58, 1.20, 1.52, 1.60, 1.55, 1.67, 
        1.77, 1.50, 1.64, 1.54, 1.40, 1.36, 1.50, 
        1.40, 1.35, 1.48, 1.64, 1.91, 1.70)

What is the true mean of $B/A$? Could it be the golden ratio?!?

Check if $B/A$ measurements come from a Normal distribution.

qqnorm(scale(gr))
abline(0,1)

Estimation

Based on $X_1,\dots ,X_n$, define the sample statistics

• ${\displaystyle {\overline{X}}_n=\frac{1}{n}\sum _{i=1}^n{X}_i}$
• ${\displaystyle S_n^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-{\overline{X}}_n{)}^2}$

Then ${\overline{X}}_n$ and $S_n^2$ are unbiased estimators of $\mu$ and ${\sigma }^2$, respectively.

Golden ratio example (cont):

We have ${\overline{X}}_n=$ mean(gr) = 1.565 and $S_n^2=$ var(gr) = 0.0219.

hist(gr)

Important sampling distribution results

Provided $X_1,\dots ,X_n\stackrel{\mathrm{ind}}{\sim }\mathcal{N}(\mu ,{\sigma }^2)$, we have

• ${\displaystyle \frac{{\overline{X}}_n-\mu }{\sigma /\sqrt{n}}\sim \mathcal{N}(0,1)}$
• ${\displaystyle \frac{(n-1)S_n^2}{{\sigma }^2}\sim {\chi }_{n-1}^2}$
• ${\displaystyle \frac{{\overline{X}}_n-\mu }{S_n/\sqrt{n}}\sim t_{n-1}}$

Discuss: Anatomy of chi-square and t random variables

• ${\displaystyle Z_1,\dots ,Z_m\stackrel{\mathrm{ind}}{\sim }\mathcal{N}(0,1)⟹Z_1^2+\cdots +Z_m^2\sim {\chi }_m^2}$.
• ${\displaystyle Z\sim \text{Normal(0,1)}\perp \perp W\sim {\chi }_m^2⟹\frac{Z}{\sqrt{W/m}}\sim t_m}$.

Relate these to the results on the previous slide.

Simulation illustrating sampling distribution results:

sims <- 1000
mu <- 1
sigma <- 1/2
n <- 8
Tn <- numeric(sims)
Wn <- numeric(sims)
for(s in 1:sims){
  
  X <- rnorm(n,mu,sigma)
  sn <- sd(X)
  xbar <- mean(X)
  Tn[s] <- sqrt(n)*(xbar - mu) / sn
  Wn[s] <- (n-1)*sn^2 / sigma^2
  
}

hist(Tn,freq = FALSE,breaks = 50)
x <- seq(-4,4,length = 500)
lines(dt(x,n-1)~x)

hist(Wn,freq = FALSE,breaks = 50)
x <- seq(0,max(Wn),length = 500)
lines(dchisq(x,n-1)~x)

Confidence intervals for the mean and variance

The sampling distribution results give $(1-\alpha )100\mathrm{\%}$ CIs as

• ${\displaystyle {\overline{X}}_n\pm t_{n-1,\alpha /2}\frac{S_n}{\sqrt{n}}}$ for $\mu$.
• ${\displaystyle (\frac{(n-1)S_n^2}{{\chi }_{n-1,\alpha /2}^2},\frac{(n-1)S_n^2}{{\chi }_{n-1,1-\alpha /2}^2})}$ for ${\sigma }^2$.

Exercise: Derive the above.

Golden ratio example (cont):

Build 95% CIs for population mean and variance of $B/A$ values:

alpha <- 0.05
n <- length(gr)

lomu <- mean(gr) - qt(1-alpha/2,n-1) * sd(gr)/sqrt(n)
upmu <- mean(gr) + qt(1-alpha/2,n-1) * sd(gr)/sqrt(n)

losgs <- (n-1) * var(gr) / qchisq(1-alpha/2,n-1)
upsgs <- (n-1) * var(gr) / qchisq(alpha/2,n-1)

The 95% CI for $\mu$ is (1.506,1.623). For ${\sigma }^2$ it is (0.014,0.041).

Testing hypotheses about the mean

Consider testing hypotheses about $\mu$ of the form $$\begin{array}{lclcl}{H}_0\text{: }\mu \ge {\mu }_0& \text{ or }& H_0\text{: }\mu ={\mu }_0& \text{ or }& H_0\text{: }\mu \le {\mu }_0\\ H_1\text{: }\mu <{\mu }_0& & H_1\text{: }\mu \ne {\mu }_0& & H_1\text{: }\mu >{\mu }_0.\end{array}$$ Reject or fail to reject $H_0$ based on the value of the test statistic $$T_{\mathrm{stat}}=\frac{{\overline{X}}_n-{\mu }_0}{S_n/\sqrt{n}}.$$ Rejection rules for the above at significance level $\alpha$ are $$\begin{array}{lclcl}{T}_{\mathrm{stat}}<-t_{n-1,\alpha }& \text{ or }& |T_{\mathrm{stat}}|>t_{n-1,\alpha /2}& \text{ or }& T_{\mathrm{stat}}>t_{n-1,\alpha }.\end{array}$$

The corresponding p-values are, with $T\sim t_{n-1}$, the probabilities $$\begin{array}{lclcl}P(T<T_{\mathrm{stat}})& \text{ or }& 2\times P(T>|T_{\mathrm{stat}}|)& \text{ or }& P(T>T_{\mathrm{stat}}).\end{array}$$

Golden ratio example (cont):

Test $H_0$: $\mu =1.618$ vs $H_1$: $\mu \ne 1.618$ at $\alpha =0.05$ based on data.

alpha <- 0.05
Tstat <- (mean(gr) - 1.618) / (sd(gr) / sqrt(n))
abs(Tstat) > qt(1-alpha/2,n-1)

[1] FALSE

Fail to reject $H_0$ since $T_{\mathrm{stat}}$ = -1.866 is smaller in absolute value than $t_{n-1,\alpha /2}$ = 2.056.

pval <- 2*(1 - pt(abs(Tstat),n-1))

Equivalently, the p-value, which is 0.073, is greater than $\alpha =0.05$.

The t.test() function in R

The function t.test() tests $H_0$: $\mu =0$ vs $H_1$: $\mu \ne 0$ by default.

t.test(gr)

    One Sample t-test

data:  gr
t = 54.902, df = 26, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 1.506228 1.623401
sample estimates:
mean of x 
 1.564815 

The t.test() function in R

Now test $H_0$: $\mu =1.618$ versus $H_1$: $\mu \ne 1.618$, ask for 99% CI.

t.test(gr, mu = 1.618, conf.level = 0.99)

    One Sample t-test

data:  gr
t = -1.866, df = 26, p-value = 0.07336
alternative hypothesis: true mean is not equal to 1.618
99 percent confidence interval:
 1.485616 1.644013
sample estimates:
mean of x 
 1.564815 

The t.test() function in R

Now test $H_0$: $\mu \le 1.618$ versus $H_1$: $\mu >1.618$.

t.test(gr, mu = 1.618, alternative = "greater")

    One Sample t-test

data:  gr
t = -1.866, df = 26, p-value = 0.9633
alternative hypothesis: true mean is greater than 1.618
95 percent confidence interval:
 1.516202      Inf
sample estimates:
mean of x 
 1.564815 

Testing hypotheses about the variance

Consider testing hypotheses about ${\sigma }^2$ of the form $$\begin{array}{lcl}{H}_0\text{: }{\sigma }^2\ge {\sigma }_0^2& \text{ or }& H_0\text{: }{\sigma }^2\le {\sigma }_0^2\\ H_1\text{: }{\sigma }^2<{\sigma }_0^2& & H_1\text{: }{\sigma }^2>{\sigma }_0^2\end{array}$$ Reject or fail to reject $H_0$ based on the value of the test statistic $$W_{\mathrm{stat}}=\frac{(n-1)S_n^2}{{\sigma }_0^2}.$$ Rejection rules for the above at significance level $\alpha$ are $$\begin{array}{lcl}{W}_{\mathrm{stat}}<{\chi }_{n-1,1-\alpha }^2& \text{ or }& W_{\mathrm{stat}}>{\chi }_{n-1,\alpha }^2\end{array}$$

The corresponding p-values are, with $W\sim {\chi }_{n-1}^2$, the probabilities $$\begin{array}{lcl}P(W<W_{\mathrm{stat}})& \text{ or }& P(W>W_{\mathrm{stat}}).\end{array}$$

Golden ratio example (cont):

Test $H_0$: ${\sigma }^2\ge 0.03$ vs $H_1$: ${\sigma }^2<0.03$ at $\alpha =0.05$ based on data.

alpha <- 0.05
Wstat <- (n-1)*var(gr) / 0.03
Wstat < qchisq(alpha,n-1)

[1] FALSE

FTR $H_0$ since $W_{\mathrm{stat}}$ = 19.009 is not less than ${\chi }_{n-1,1-\alpha }^2$ = 15.379.

pval <- pchisq(Wstat,n-1)

Equivalently, the p-value, which is 0.164, is greater than $\alpha =0.05$.

Pallet weights example

The weights in lbs of several wooden pallets were recorded.

wt <- c(41,34,40,44,33,42,52,38,32,31,31,35,
        39,44,42,42,35,33,40,48,51,32,41,35,
        38,48,37,35,42,41,40,47,40,46,33,38,
        51,39,40)

Pallet weights example continued

• Check if we can assume normally distributed pallet weights.
• Build a 95% CI for the mean pallet weight assuming $\sigma =5$.
• Build a 95% CI for the mean pallet weight assuming $\sigma$ is unknown.
• Test at $\alpha =0.05$ whether the mean pallet weight is equal to $48$ lbs.
• Test at $\alpha =0.05$ whether the mean pallet weight is less than $40$ lbs.

Sample size calculations

We can choose a sample size based on the desired:

1. Width of a confidence interval.
2. Power of a test to reject $H_0$ when it is false.

Sample size required to achieve desired CI width

A CI for $\mu$ takes the form ${\overline{X}}_n\pm M$, where

• ${\displaystyle M=z_{\alpha /2}\frac{\sigma }{\sqrt{n}}}$ if $\sigma$ is known
• ${\displaystyle M=t_{n-1,\alpha /2}\frac{S_n}{\sqrt{n}}}$ if $\sigma$ is unknown

For ease, use the “$\sigma$-known” version.

If one wants $M\le M^{\ast }$, find smallest $n$ such that ${\displaystyle z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le M^{\ast }}$.

So take ${\displaystyle n=\lceil {\left(\frac{z_{\alpha /2}\sigma }{M^{\ast }}\right)}^2\rceil }$, where $\lceil \cdot \rceil$ rounds up.

Must put in a guess for $\sigma$.

Golden ratio example (cont):

Find $n$ required to make the 95% CI for $\mu$ no wider than $0.08$.

alpha <- 0.05
M <- 0.08/2
sigma_guess <- sd(gr)
nr <- ceiling((qnorm(1-alpha/2) * sigma_guess / M)^2)
nr

[1] 53

Sample size required to achieve desired power

The power of a test is the probability with which it rejects $H_0$.

For tests of $H_0$ concerning the mean $\mu$ we write the power as $$\gamma (\mu )=P(\text{Reject }{H}_0\text{ when true mean is }\mu )=P_{\mu }(\text{Reject }{H}_0).$$

So the power depends on the true value of $\mu$, i.e. is a function of $\mu$.

Exercise: Derive the power functions for the tests of $$\begin{array}{lclcl}{H}_0\text{: }\mu \ge {\mu }_0& \text{ and }& H_0\text{: }\mu ={\mu }_0& \text{ and }& H_0\text{: }\mu \le {\mu }_0\\ H_1\text{: }\mu <{\mu }_0& & H_1\text{: }\mu \ne {\mu }_0& & H_1\text{: }\mu >{\mu }_0\end{array}$$ with the rejection rules $$\begin{array}{lclcl}{Z}_{\mathrm{stat}}<-z_{\alpha }& \text{ and }& |Z_{\mathrm{stat}}|>z_{\alpha /2}& \text{ and }& Z_{\mathrm{stat}}>z_{\alpha },\end{array}$$ respectively, where ${\displaystyle Z_{\mathrm{stat}}=\frac{{\overline{X}}_n-{\mu }_0}{\sigma /\sqrt{n}}}$ ($\sigma$-known case).

Plot of power curves for right-, left-, and two-sided tests

Set ${\mu }_0=1.618$, $n=27$, and $\alpha =0.05$ and use $\sigma =0.148$.

alpha <- 0.05
mu0 <- 1.618
sigma <- sd(gr)
n <- length(gr)
mu <- seq(mu0-4*sigma/sqrt(n),mu0+4*sigma/sqrt(n),length=500)
za <- qnorm(1-alpha)
za2 <- qnorm(1-alpha/2)
d <- sqrt(n) * (mu - mu0) / sigma
rp <- 1 - pnorm(za - d)
lp <- pnorm(-za - d)
rp2 <- 1 - pnorm(za2 - d)
lp2 <- pnorm(-za2 - d)
tsp <- lp2 + rp2

plot(rp ~ mu, type = "l", ylab = "power", xlab = "mu")
lines(lp ~ mu, lty =2)
lines(tsp ~ mu, lty = 3)

Power curve for right-sided test at various sample sizes

Test $H_0$: $\mu \le 1.618$ vs $H_1$: $\mu >1.618$. Use $\alpha =0.05$ and $\sigma =0.148$.

alpha <- 0.05
mu0 <- 1.618
sigma <- sd(gr)
n <- length(gr)
mu <- seq(mu0-1*sigma/sqrt(n),mu0+4*sigma/sqrt(n),length=500)
za <- qnorm(1-alpha)

# various sample sizes
nn <- c(10,20,30,40,50,60)
rp <- matrix(NA,500,length(nn))
for(j in 1:length(nn)){
  d <- sqrt(nn[j]) * (mu - mu0) / sigma
  rp[,j] <- 1 - pnorm(za - d)
}

plot(NA,xlim = range(mu), ylim = c(0,1), ylab = "power", xlab = "mu")
for(j in 1:length(nn)) lines(rp[,j] ~ mu, lty = j)
legend(x = min(mu), y = 1,legend = paste("n =",nn),lty = 1:length(nn),bty = "n")

Sample size based on desired power

To find the smallest sample size guaranteeing a desired power:

1. Fix an alternative value ${\mu }^{\ast }$ and a desired power ${\gamma }^{\ast }$.
2. Set up the equation $\gamma ({\mu }^{\ast })={\gamma }^{\ast }$ and solve for $n$ (then round up).

For our tests concerning $\mu$ when $\sigma$ is known, we obtain:

• In the one-sided case ${\displaystyle n=\lceil {\sigma }^2(\frac{z_{\alpha }+z_{{\beta }^{\ast }}}{{\mu }^{\ast }-{\mu }_0}{)}^2\rceil }$.
• In the two-sided case ${\displaystyle n=\lceil {\sigma }^2(\frac{z_{\alpha /2}+z_{{\beta }^{\ast }}}{{\mu }^{\ast }-{\mu }_0}{)}^2\rceil }$.

In the above ${\beta }^{\ast }=1-{\gamma }^{\ast }$.

Exercise: Derive the sample size formula for the test of $$H_0\text{: }\mu \le {\mu }_0\text{ vs }{H}_1\text{: }\mu >{\mu }_0$$ when $\sigma$ is known.

Golden ratio example (cont):

Suppose the true mean of $B/A$ in the population is $1.65$.

Give the sample size $n$ required to reject $H_0$: $\mu \le 1.618$ vs $H_1$: $\mu >1.618$ with power $\ge 0.80$. Use $S_n$ = 0.148 as a guess of $\sigma$.

alpha <- 0.05
gm <- 0.80
sigma <- sd(gr)
mu <- 1.65
mu0 <- 1.618
za <- qnorm(1 - alpha)
zb <- qnorm(gm)
nr <- ceiling(sigma^2 * (za + zb)^2 / (mu - mu0)^2)
nr

[1] 133