# Confidence intervals continued

04/12/2021

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## Outline

• The following topics will be covered in this lecture:

• A review of confidence intervals
• Estimating the necessary sample size for an error tolerance
• One-sided confidence bounds
• Confidence intervals for the mean, with unknown population standard deviation
• Large sample size
• Student t distribution
• Student t confidence intervals

### Introduction to confidence intervals – continued

Courtesy of Mario Triola, Essentials of Statistics, 6th edition

• Let’s take an example confidence level of $$95\%$$ – this corresponds to a rate of failure of $$5\%$$ over infinitely many replications.
• Generally, we will write the confidence level as, $(1 - \alpha) \times 100\%$ so that we can associate this confidence level with its rate of failure $$\alpha$$.
• In our problem, because $$Z = \frac{\overline{X} - \mu }{\frac{\sigma}{\sqrt{n}}}$$ has a standard normal distribution, we may write \begin{align} P\left(-z_\frac{\alpha}{2} \leq \frac{\overline{X} - \mu }{\frac{\sigma}{\sqrt{n}}} \leq z_\frac{\alpha}{2}\right) = 1 - \alpha \end{align} where
• we want to find the critical value $$z_\frac{\alpha}{2}$$ for which:
• $$(1-\frac{\alpha}{2})\times 100\%$$ of the area under the normal density lies to the left of $$z_\frac{\alpha}{2}$$; and
• $$(1-\frac{\alpha}{2})\times 100\%$$ of the area under the normal density lies to the right of $$-z_\frac{\alpha}{2}$$.
• Put together, $$(1-\alpha)\times 100\%$$ of values lie within $$[-z_\frac{\alpha}{2},z_\frac{\alpha}{2}]$$ in the standard normal.

### Introduction to confidence intervals – continued

Courtesy of Mario Triola, Essentials of Statistics, 6th edition

• In the figure to the left, we see exactly thus how to find the width of the region for a given confidence level.
• For a given confidence level $$(1 -\alpha)\times 100\%$$, we will find the particular $$\alpha$$.
• We then find the associated $$z_\frac{\alpha}{2}$$ critical value.
• This critical value is associated to the measure of extremeness of finding an observation that lies far away from the mean.
• Particularly, only $$\alpha \times 100\%$$ of the population lies outside of the region $$[-z_\frac{\alpha}{2},z_\frac{\alpha}{2}]$$ in the standard normal.
• Manipulating the earlier probability statement, we find that \begin{align} & P\left(-z_\frac{\alpha}{2} \leq \frac{\overline{X} - \mu }{\frac{\sigma}{\sqrt{n}}} \leq z_\frac{\alpha}{2}\right) = 1 - \alpha\\ \Leftrightarrow & P\left( \overline{X} - \frac{\sigma}{\sqrt{n}}z_\frac{\alpha}{2} \leq \mu \leq \overline{X} + \frac{\sigma}{\sqrt{n}} z_\frac{\alpha}{2}\right) = 1 - \alpha. \end{align}
• The above represents the traditional way to construct a confidence interval; with $$(1-\alpha)\times 100\%$$ confidence $L \leq \mu \leq U \Leftrightarrow \overline{X} - \frac{\sigma}{\sqrt{n}}z_\frac{\alpha}{2} \leq \mu \leq \overline{X} + \frac{\sigma}{\sqrt{n}} z_\frac{\alpha}{2}$

### Introduction to confidence intervals – continued

• Note, the last argument only works for a normal population.
• However, we get a very good approximation with $$n$$ large enough for non-normal populations.
• The last argument also required that the variance $$\sigma^2$$ was known.

• If $$\sigma^2$$ is unknown, we need to change the argument slightly, which is the reason we belabored z-scores in this argument.
• To compute the confidence intervals as above, we need to introduce a new function, the quantile function:

• qnorm(p, mean, sd) – this is the quantile function that gives the critical value associated to the value $$\alpha=p$$.
• The quantile function returns the value $$z_p$$ for which $$p\times 100 \%$$ of the area under the standard normal lies to the left of $$z_p$$ and $$(1-p)\times 100\%$$ lies to the right.
• Compared to our earlier example:
qnorm(0.025)

[1] -1.959964

qnorm(0.975)

[1] 1.959964

• By the symmetry of the normal, we can thus compute $$z_\frac{\alpha}{2}$$=qnorm(p) where $$p=1-\frac{\alpha}{2}$$.

### Introduction to confidence intervals – continued

• Suppose we know that, $$\overline{X}$$ is the sample mean from a normal population with (unknown) mean $$\mu= 10$$, standard deviation $$\sigma= 2$$ and sample size $$n=16$$, then

$\overline{X} \sim N\left(10, \frac{4}{16}\right).$

• Notice that the standard error is thus given as $$\sigma_\overline{X} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$

• If $$\overline{X}$$ is observed to take the value $$\overline{x}=9$$, then we can construct the $$95\%$$ confidence interval for $$\mu$$ as,

se <- 0.5
z_alpha_over_2 <- qnorm(0.975)
ci <- c(9 - se*z_alpha_over_2 , 9+se*z_alpha_over_2)
ci

[1] 8.020018 9.979982

• Notice, the above confidence interval does not contain the true population mean $$\mu=10$$…

### Introduction to confidence intervals – continued

• Let's suppose instead we repeat the argument but select $$\alpha=0.01$$.

• This corresponds to a critical value $$z_\frac{\alpha}{2}=z_{0.005}$$.

• To compute the the corresponding critical value, we are looking for p=$$1-\frac{\alpha}{2}=0.995$$

se <- 0.5
z_alpha_over_2 <- qnorm(0.995)
ci <- c(9 - se*z_alpha_over_2 , 9+se*z_alpha_over_2)
ci

[1]  7.712085 10.287915

• Notice that this $$(1-\alpha)\times 100\% = 99\%$$ confidence interval does contain the true mean.

• This confidence interval is somewhat wider as we want to guarantee that the procedure will work $$99\%$$ of the time.

• Therefore, we need to include a wider range of plausible values when we construct such an interval.

### Introduction to confidence intervals – continued

• What we are imagining when we construct confidence intervals is the following.
• Based on some particular sample $$X_{j,1},\cdots, X_{j,n}$$ of size $$n$$ indexed by $$j$$, we will get some particular value for the confidence interval.
• If we replicate the sample of size $$n$$, indexed by $$j$$, we will almost surely find a new confidence interval based on each replicate.

Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• Our goal in constructing a confidence interval is thus to catch the true parameter value with the confidence level $$(1-\alpha)\times 100\%$$ out of all replicates.
• If we want higher confidence, we need wider intervals to catch the true value at a higher rate.
• However, the normal confidence interval, $\overline{X} - \frac{\sigma}{\sqrt{n}}z_\frac{\alpha}{2} \leq \mu \leq \overline{X} + \frac{\sigma}{\sqrt{n}} z_\frac{\alpha}{2}$ also has a width that depends on the sample size.
• This is of course, as we discussed in the central limit theorem, the precision of the sample mean $$\overline{X}$$ increases for larger sample sizes, with a standard deviation that shrinks like $$\frac{1}{\sqrt{n}}$$.
• This allows us to select a sample size for a target precision, given a level of confidence.

## Choosing the sample size

• In situations whose sample size can be controlled, we can choose $$n$$ so that we are $$(1 − \alpha)\times 100\%$$ confident that the error in estimating $$\mu$$ is less than a specified bound on the error $$E$$.
• The appropriate sample size is found by choosing $$n$$ such that $z_\frac{\alpha}{2} \hat{\sigma}_\overline{X} = z_\frac{\alpha}{2} \frac{\sigma}{\sqrt{n}} = E.$

Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• Notice, $$n$$ does not affect the choice of $$z_\frac{\alpha}{2}$$ which is based on the desired confidence level;
• rather, this changes the size of the standard error, which decreases proportionally to $$\frac{1}{\sqrt{n}}$$.
• Solving this equation gives the following formula for $$n$$: \begin{align} &E = z_\frac{\alpha}{2} \frac{\sigma}{\sqrt{n}} \\ \Leftrightarrow & \frac{1}{\sqrt{n}} = \frac{E}{z_\frac{\alpha}{2} \sigma} \\ \Leftrightarrow & n = \left(\frac{z_\frac{\alpha}{2} \sigma}{E}\right)^2 \end{align}
• For a specified error tolerance $$E$$, we can thus choose $$n$$ such that $$\mu$$ lies within $$E = z_\frac{\alpha}{2} \frac{\sigma}{\sqrt{n}}$$ of the sample mean $$\overline{X}$$ with $$(1-\alpha)\times 100 \%$$ confidence.
• $$E$$ is a chosen radius of the confidence interval around the point estimate, in which we want to “catch” the true value.

### Choosing the sample size – continued

• Putting this formally:
Sample Size for Specified Error on the Mean, Variance Known
If $$\overline{X}$$ is used as an estimate of $$\mu$$, we can be $$(1 − \alpha)\times 100\%$$ confident that the error $$\vert \overline{x} - \mu\vert$$ will not exceed a specified amount $$E$$ when the sample size is $n = \left(\frac{z_\frac{\alpha}{2} \sigma}{E}\right)^2.$
• Note that the above was directly a consequence of the central limit theorem, and the fact that

$\overline{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right).$

when the underlying population is normal (or approximately for large $$n$$ for non-normal populations).

### Choosing the sample size – example

• Suppose that we want to determine how many specimens must be tested to ensure that the $$95\%$$ CI on $$\mu$$ for a normal population with known standard deviation $$\sigma=1$$ falls below some specified error tolerance.

• For a $$95\%$$ confidence interval, we have $$\alpha=1-0.95=0.05$$.

• We therefore require that $$1-\frac{\alpha}{2} = 1-0.025=0.975$$ of the area lies to the left and $$\frac{\alpha}{2}=0.025$$ lies to the right of $$z_\frac{\alpha}{2}$$.

• If $$E=10\%$$, we would write,

\begin{align} n &= \left(\frac{z_\frac{\alpha}{2} \sigma}{E}\right)^2\\ &= \left(\frac{z_{0.025} \times 1}{0.10}\right)^2 \end{align}

• In R we can write

E <- 0.1
sigma <- 1.0
z_alpha_over_2 <- qnorm(0.975)
n_ten_percent <- ( (z_alpha_over_2 * sigma) / E )^2
n_ten_percent

[1] 384.1459


### Choosing the sample size – example

• Notice, with our last choice of a $$10\%$$ error tolerance we had a corresponding sample size of
n_ten_percent

[1] 384.1459

• Suppose we require this estimate to be more precise, say $$E=5\%$$ or $$E=1\%$$.

• We can revise our previous estimate by

E <- 0.05
n_five_percent <- ( (z_alpha_over_2 * sigma) / E )^2
n_five_percent

[1] 1536.584

• and
E <- 0.01
n_one_percent <- ( (z_alpha_over_2 * sigma) / E )^2
n_one_percent

[1] 38414.59


### Choosing the sample size – example

• In the last example, we saw the necessary sample size grow as follows:
x_vals <- c(0.01, 0.05, 0.10)
y_vals <- c(n_one_percent, n_five_percent, n_ten_percent)
par(cex = 2.0, mar = c(5, 4, 4, 2) + 0.3)
plot(x_vals, y_vals, xlab = "Error tolerance", ylab="Necessary sample size", type="b")