04/12/2021

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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

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.

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**\[ L \leq \mu \leq U \Leftrightarrow \overline{X} - \frac{\sigma}{\sqrt{n}}z_\frac{\alpha}{2} \leq \mu \leq \overline{X} + \frac{\sigma}{\sqrt{n}} z_\frac{\alpha}{2} \]

- 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.

- However, we get a
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 \( p=1-\frac{\alpha}{2} \).

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 \( \mu=10 \)…

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.

- 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.

- 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.

- 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).

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
```

- 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
```

- 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")
```