Lec 21

Review

Uncertainty and Statistics

“All models are wrong, but some are useful” - George E. P. Box

What is the primary goal of statistics?

Can we achieve this goal without making any errors?

We define uncertainty from two central concepts:

For every population we can draw information from, there is a sample \(X\)

The mean of our sample is \(\bar{x}\)

\(\bar{x}\) is assumed to be normally distributed under two circumstances

The distribution of \(X\), and thus our population, is NORMAL

Our sample size of \(X\) is \(>30\)

Why does point (1) make sense?

If we draw samples from a population that has any distribution

It’s samples will have the same distribution

A condition where this doesn’t hold is when we transform our data somehow
- i.e., Taking samples from a normally distributed population but transforming the samples to strictly positive, discrete values

Why does point (2) make sense?

In applied statistics we check our assumptions

Logically the only way to check this assumption would be to take a lot of samples

We’ve done that outside of this class, our assumptions do hold

We assume that the distribution of \(\bar{x}\) is normal even upon one observation arising from a sample of \(n>30\)

Empirical and mathematical proof are shockingly easy to obtain (and do by hand!)

I do recommend just seeking out a program to do it for you however

When we sample from a population, and compute a statistic, we’re always slightly wrong

Statistics accepts errors on the single condition that you can quantify those errors

Central Limit Theorem

Given we have an understanding of the CLT, most of this class should feel essentially intuitive

“…a large, properly draw sample will resemble the population from which it is drawn.” - Charles Wheelan

Think about sampling \(40\) defensive linemen from random college football teams

We can pretty intuitively assume that they’re going to be pretty heavy on average

If we got an average weight of \(140\) lbs. we’d be concerned we sampled wrong

Not because it’s impossible

But because it’s highly unlikely

That’s it.

Confidence Intervals are a way to quantify uncertainty

Put simply, we’re saying “The true value is \(X\), give or take \(a\)”

We have two methods for calculating confidence intervals:

\(z\)-method (given \(\sigma\) is known)

\[\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\]

\(t\)-method (given \(\sigma\) is unknown)

\[\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]

\(t\)-method refers directly to the Students t-distribution

Similar to the standard normal distribution:

Unimodal
Symmetric around \(0\)

Dissimilar from the standard normal distribution:

Wider (or heavier) tails than the standard normal
- Meaning it’s more spread out

Distinguished by degrees of freedom (\(df = n - 1\))
- As \(df\) increase the t-distribution converges to a normal distribution

In practice, we always use the t-method

Why?

Questions?

Goals for today:

Develop a method for dealing with uncertainty in proportional data
Introduce the fundamental concepts of Hypothesis testing

Uncertainty II

Confidence Intervals for Proportions

To get a confidence interval we need:

Point estimate
Margin of Error

Our point estimate for \(p\) (population proportion):

\[\hat{p}\]

Margin of error is given by:

\[ z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

We need to justify being able to make a confidence interval

Given the CLT for proportions:

\[ n\hat{p} \geq 10 \quad \text{and} \quad n(1 - \hat{p}) \geq 10 \]

then the \(100(1 - \alpha)\%\) confidence interval for \(p\) is:

\[ \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \quad \text{(Margin of Error)} \]

Recall that the critical value \(z_{\alpha/2}\) comes from the standard normal distribution.

Sleep apnea is a disorder in which there are pauses in breathing during sleep. In a sample of 427 people aged 65 or over, 104 had sleep apnea.

a. Find a point estimate for the population proportion of those aged 65 and over who have sleep apnea

\[ \hat{p} = \frac{104}{427} \approx 0.244 \]

b. Compute the margin of error for a 99% confidence level

For 99% confidence, \(\alpha/2 = 0.005 \Rightarrow z_{\alpha/2} = 2.576\). The margin of error is:

\[ z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 2.576 \times \sqrt{\frac{0.244(1 - 0.244)}{427}} \approx 0.054 \]

c. Construct a 99% confidence interval for the proportion of those aged 65 or over who have sleep apnea

We construct the 99% confidence interval for \(p\) by:

\[ \hat{p} \pm \text{Margin of Error} \Rightarrow 0.244 \pm 0.054 \]

or \((0.190, 0.298)\)

d. Does it appear that more than 9% of elderly people have sleep apnea? Yes, all the values in the CI are greater than 0.09.

Researchers want to estimate the mean amount of sleep per night for students at Upper Midwest University. A sample of 15 students had an average of 6.4 hours per night and a standard deviation of 1.27 hours.

a. Should we use the z-method or t-method?

\(t\) method, because \(\sigma\) is unknown

b. Find a point estimate for the mean amount of sleep per night for UMU students

\[ \bar{x} = 6.4 \]

c. Compute the margin of error for a 90% confidence level

\[ \alpha = 1 - 0.90 = 0.10 \Rightarrow \alpha/2 = 0.05 \]

With \(df = n - 1 = 14\), \(t_{\alpha/2} = 1.761\) from the t-table. Compute the margin of error:

\[ t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.761 \times \frac{1.27}{\sqrt{15}} \approx 0.577 \]

d. Construct a 90% confidence interval for the mean amount of sleep per night for students at UMU

The 90% CI for the mean amount of sleep is:

\[ \bar{x} \pm \text{Margin of Error} \Rightarrow 6.4 \pm 0.577 \]

\[\text{or } (5.823, 6.977)\]

e. Does the confidence interval contradict a claim that the mean amount of sleep for UMU students is 6 hours? Explain

No, because 6 is contained in the confidence interval

A college admissions officer sampled 107 freshmen and found that 38 scored more than 510 on the math SAT.

a. Find a point estimate for the proportion of all entering freshmen at this college who scored more than 510 on the math SAT

\[ \hat{p} = \frac{38}{107} \approx 0.355 \]

b. Compute the margin of error for a 95% confidence level

\[ \alpha = 1 - 0.95 = 0.05 \Rightarrow \alpha/2 = 0.025 \Rightarrow z_{\alpha/2} = 1.96 \]

Compute the margin of error:

\[ z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 1.96 \times \sqrt{\frac{0.355(1 - 0.355)}{107}} \approx 0.091 \]

c. Construct a 95% confidence interval for the proportion of all entering freshmen who scored more than 510 on the math SAT

\[ \hat{p} \pm \text{Margin of Error} \Rightarrow 0.355 \pm 0.091 \]

\[\text{or } (0.264, 0.446)\]

We are 95% confident that between 26.4% and 44.6% of the entering freshmen scored more than 510 on the math SAT

Hypothesis Testing

My philosophy for scientific communication should be apparent by now:

Why should you care?
What does this do?
Why do we use this?

Making claims about statements is how science works

We can convert statements into parameters, and make claims about them using statistics

We call this: “Hypothesis Testing”

In hypothesis testing, there are two competing statements about population parameters:

\[H_0\equiv \text{null hypothesis} \quad \text{vs} \quad H_1 \equiv \text{alternate hypothesis}\]

Upon forming these statements:

We use data collected from a sample to test if we
- Reject \(H_0\), thereby supporting \(H_1\)

A statistical hypothesis is often a statement about population parameter(s)

The null hypothesis, \(H_0\), states that the parameter is equal to a specific value

\[H_0 : \mu = 35\]

The alternate hypothesis, \(H_1\), states that the value of the parameter differs from the value specified by the null hypothesis

\[H_1 : \mu < 35\]

\[H_1 : \mu > 35\]

\[H_1 : \mu \neq 35\]

There are three types of alternate hypothesis

Consider \(H_0 : \mu = 35\)

\(H_1 : \mu < 35\) \(\Rightarrow\) called left-tailed alternate hypothesis
\(H_1 : \mu > 35\) \(\Rightarrow\) called right-tailed alternate hypothesis
\(H_1 : \mu \neq 35\) \(\Rightarrow\) called two-tailed alternate hypothesis

Left-tailed and right-tailed hypotheses are called one-tailed hypotheses

A null hypothesis is generally thought of as a default state of nature (e.g. existing knowledge)

An alternate hypothesis, on the other hand, contradicts the default state (e.g. new knowledge)

In most cases, whatever we wish to establish is placed in the alternate hypothesis

After developing \(H_0\) and \(H_1\), we collect a set of data

Based on the data, we construct a test statistic to reach one of the following decisions:
- Reject \(H_0\)
- Fail to reject \(H_0\)

If we reject \(H_0\)

We conclude that \(H_1\) is true

If we fail to reject \(H_0\)

We conclude that the data do not provide enough evidence to reject \(H_0\)

Errors in Hypothesis Testing

We’re making inference about populations from samples

We will be a certain amount of wrong
Depending on how wrong we are, and in what way, we define two types of errors

Type I error: \(H_0\) is true in reality, but we reject \(H_0\)

Type II error: \(H_1\) is true in reality, but we do not reject \(H_0\)

\[ \begin{array}{|c|c|c|} \hline \text{Decision} & H_0 \ \text{True} & H_0 \ \text{False} \\ \hline \text{Reject} \ H_0 & \text{Type I error} & \text{Correct decision} \\ \hline \text{Don’t reject} \ H_0 & \text{Correct decision} & \text{Type II error} \\ \hline \end{array} \]

The probability of having the Type I error is denoted by \(\alpha\)

The probability of having the Type II error is denoted by \(\beta\)

Minimizing both \(\alpha\) and \(\beta\) is impossible due to a trade-off between \(\alpha\) and \(\beta\)

If we decrease \(\alpha\), then \(\beta\) tends to increase

In general, controlling \(\alpha\)-level (chance of making Type I error) is more important since Type I error leads to the acceptance of incorrect new knowledge