[.3Cm] Part IV: Statistical Inference

Introduction to Statistics Part IV: Statistical inference

Achim Ahrens Anna Babloyan [email protected] [email protected] Erkal Ersoy [email protected]

Heriot-Watt University, Edinburgh

September 2015 Outline 1. Descriptive statistics

I Sample statistics (mean, variance, percentiles) I Graphs (box plot, histogram) I Data transformations (log transformation, unit of measure) I Correlation vs. Causation 2. Probability theory

I Conditional probabilities and independence I Bayes’ theorem 3. Probability distributions

I Discrete and continuous probability functions I Probability density function & cumulative distribution function I Binomial, Poisson and Normal distribution I E[X] and V[X] 4. Statistical inference

I Population vs. sample I Law of large numbers I Central limit theorem I Conﬁdence intervals I Hypothesis testing and p-values 1 / 41 Introduction Recall, in the last lecture we assumed that we know the probability distribution of the random variable in question as well as the parameters of the distribution (e.g. µ and σ2 for the normal distribution). Under these assumptions we were able to obtain the probability that the random variable would take values within a particular interval (e.g. P(X≤8)).

0.5 N(0, 1) N(0, 2) 0.4 N(0, 3)

0.3 ) x ( f 0.2

0.1

0 8 6 4 2 0 2 4 6 8 − − − − x What if we don’t know µ? 2 / 41 Population vs. sample Suppose we are interested in the distribution of heights in the UK. The residents of the UK are the population; the parameter µ is the true average height of UK residents and σ2 the true variance. If we were to measure the height of all UK residents, we would conduct a census. However, measuring the height of every individual is hardly feasible, or only at an exorbitant cost. Instead, we can randomly select a sample from the population and make inferences from the sample to the population. In particular, we can use the sample statistics (e.g. sample mean and sample variance) to make inferences about the true, but unknown population parameters (µ and σ2).

3 / 41 Population vs. sample We randomly select a sample from the UK population and measure the heights of the individuals in the sample. Simple random sample A random sample is given if each individual in the population has an equal chance of being chosen.

Since the draws are random, the height of the ﬁrst, second, third, . . . nth selected individual is random, too. That is, X1, X2,..., Xn are random variables. I.I.D.

Suppose we draw n items (X1, X2,..., Xn) at random from the same population. Since X1, X2,..., Xn are drawn from the same population, they are identically distributed. Furthermore, since the distribution of Xi does not depend on the distribution of Xj (for i, j = 1,..., n; i 6= j), we can say that they are independently distributed. We say that X1, X2,..., Xn are independently and identically distributed (i.i.d.).

4 / 41 Population vs. sample Now, we draw (n=10, in cm) 182 197 183 171 171 162 152 157 192 174 Given this sample, what is our best guess about µ? It’s just the sample mean. n X 1 x¯ = x = (182 + ··· + 174) = 174.1 i 10 i=1 The sample mean is an unbiased and consistent estimator of the unknown population mean µ. Unbiasedness vs. consistency To understand unbiasedness, note that the sampling distribution of x¯ is centered at µ. When we repeatedly sample (more on this in a bit), x¯ is sometimes above the true value of the parameter µ and sometimes below it. However, the key aspect here is that there is no systematic tendency to overestimate or underestimate the true parameter. This makes x¯ an unbiased estimator of the parameter µ.

Unbiasedness vs. consistency An estimator is consistent if, as the sample size increases, the estimator 5 / 41 converges to the true population parameter. The Law of Large Numbers Although x¯ is rarely exactly right and varies from sample to sample, it is still a reasonable (and in fact, the best) estimate of the population mean, µ. This is because it is guaranteed to get closer to the population parameter µ as the sample size increases. Therefore, we know that if we could keep taking measurements from more subjects, eventually we would estimate the true population mean very accurately. This fact is usually referred to as the law of large numbers. It is a remarkable fact because it holds for any population. Law of Large Numbers If we randomly draw independent observations from any population with ﬁnite mean µ, the sample mean, x¯, of the observed values approaches the true mean, µ, of the population as the number of observations, n, goes to ∞.

6 / 41 LLN in Action 101 100 99 Mean of first n observations 98 97 1 5 10 50 100 500 1000 5000 10000 Number of observations, n

In the diagram on the previous slide (reproduced below), we have plotted the mean of the first n observations in our (artificially 7 / 41 generated) data set of 10,000 observations. More specifically, we generated a normally distributed variable with mean, µ, 100 and standard deviation, σ, 4. Then, to obtain each plotted point, we calculated the mean of the generated figures up to each n. Population vs. sample

The sample mean estimator X¯ is a function of X1,..., Xn,

n 1 X X¯ = X . N i i=1 Therefore, it is a random variable, whereas the sample mean of our sample, x¯ = 174.1, is a realisation. Estimator vs. estimate An estimator is a function of random variables which represent draws from a population. Thus, the estimator is a random variable. The estimator works like a method or formula for "guessing" population parameters. An estimate, on the other hand, is the numerical value that you obtain from a speciﬁc sample. An estimate is not a random variable, it’s just a number.

As any other random variable, X¯ follows a distribution. What does the distribution look like? To answer this question, we consider one of the most remarkable theorems in statistics, the central limit theorem.

8 / 41 Central limit theorem Let’s demonstrate the CLT using a simulation. We assume that Xi ∼ i.i.d. uniform(160, 180). That is, we assume that the Xi’s are uniformly distributed within the interval [160, 180]. We proceed as follows:

1. We draw n = 50 observations (x1,..., x50) from our “population” (in this case: from our uniform distribution). 2. We obtain and write down the sample mean (i.e. x¯). 3. Repeat step (1) and (2) 10,000 times. This gives us 10,000 sample means (x¯(1), x¯(2),..., x¯(10,000)). This large set of sample means should give us an idea how the theoretical distribution of X¯ looks like.

9 / 41 CLT in Action

100 repetitions .6 .5 .4 .3 Density .2 .1 0 166 168 170 172 174 Sample mean

10 / 41 CLT in Action

5000 repetitions .6 .5 .4 .3 Density .2 .1 0 166 168 170 172 174 Sample mean

10 / 41 CLT in Action

10000 repetitions .6 .5 .4 .3 Density .2 .1 0 166 168 170 172 174 Sample mean

10 / 41 Central limit theorem

10000 repetitions .6 .5 .4 .3 Density .2 .1 0 166 168 170 172 174 Sample mean The sample mean of x¯(1), x¯(2),..., x¯(10,000) is 170.0007 and the standard deviation is 0.8139225.

11 / 41 The mean and the standard deviation of x¯

If x¯ is the mean of a sample of size n, which are drawn randomly from a large sample with mean µ and standard deviation σ, then the mean of √σ the sampling distribution of x¯ is µ and its standard deviation is n . More formally, the central limit theorem can be described as follows.

Central limit theorem Suppose you draw n random numbers from an arbitrary (discrete or continuous) distribution with mean µ and variance σ2. If n is suﬃciently large, then σ2 X¯ ∼ N µ, n ¯ 1 Pn where X = n i=1 Xi.

12 / 41 Short digression: The expected value of X¯

N 1 1 1 1 1 X¯ = X X = X + X + X + ··· + X N i N 1 N 2 N 3 N N i=1 From last lecture, we know that the expectation of a sum is the sum of the expectations and thus:

N 1 1 1 1 (X¯ ) = [X X ] = [ X ] + [ X ] + ··· + [ X ] E E N i E N 1 E N 2 E N N i=1 1 1 1 = [X ] + [X ] + ··· + [X ] N E 1 N E 2 N E N 1 1 1 = µ + µ + ··· + µ N N N = µ

13 / 41 Short digression: The variance of X¯

N 1 1 1 1 1 [X¯ ] = [X X ] = [ X + X + X + ··· + X ] V V N i V N 1 N 2 N 3 N N i=1 1 1 1 = [Y ] + [Y ] + ··· + [Y ] N 2 V 1 N 2 V 2 N 2 V N 1 1 1 = σ2 + σ2 + ··· + σ2 N 2 N 2 N 2 σ2 = N

This result tells us that the sample variance decreases as the sample size increases.

14 / 41 Making statistical inferences Conﬁdence intervals Recall the following diagram from the ﬁrst lecture, where we indicated that 95% of the values in a given data set tends to lie within 2 standard deviations from the mean (for normally distributed variables).

68%

mean

mean - one SD mean + one SD

95%

mean

mean - two SDs mean + two SDs

We can use this observation to make statistical inferences. 15 / 41 Making statistical inferences Conﬁdence intervals

0.5 N(0, 1)

0.4

0.3 ) x ( f 0.2

0.1

0 4 2 0 2 4 − − x

16 / 41 Making statistical inferences Confidence intervals As discussed earlier, the sample mean x¯ is an appropriate estimator of the unknown population mean µ because it is an unbiased estimator of µ, and it approaches the true population parameter as sample size increases. We have also mentioned, however, that this estimate varies from sample to sample. So, how reliable is this estimator? To answer this question, we need to consider the spread as well. From the central limit theorem (CLT), we know that if the population mean is µ and the standard deviation is σ, then repeated samples of n observations should yield a 2 ¯ σ sample mean x¯ with the following distribution: X ∼ N µ, n . Confidence Interval A confidence interval with confidence level C consists of two parts: 1. An interval obtained from the data in the form estimate ± margin of error 2. A chosen confidence level, C, which gives the probability that the calculated interval will contain the true parameter value.

17 / 41 Conﬁdence Intervals Calculating the interval

One of the most popular values for C is 95%, which (obviously) leads to a 95% conﬁdence interval—meaning there’s 95% probability that the true population parameter lies within that conﬁdence interval (CI).

95% Conﬁdence Interval To get a 95% CI for a population mean (µ), we simply need to do: σ x¯ ± z × √ n where z is the critical value with area C between −z and z under the standard normal curve. The right part of the expression (i.e. √σ z × n ) is the margin of error.

18 / 41 Conﬁdence Intervals Calculating the interval

Example Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% confidence level?

In other words, the student wishes to estimate the true mean boiling 19 / 41 temperature of the liquid using the results of his measurements. If the measurements follow a normal distribution, then the sample mean will σ2 have the distribution N (µ, n ). Since the sample size is 6, the standard deviation of the sample mean is equal to √1.2 = 0.49. 6 Then, the 95% CI is simply 101.82 ± 1.96 × 0.49 = [100.86, 102.78]

If we wanted to construct a 99% CI, however, our critical values would change. This would also change our CI itself:

101.82 ± 2.56 × 0.49 = [100.57, 103.07] Confidence Intervals Behaviour of confidence intervals Confidence intervals get smaller as: 1. The number of observations, n, increases

2. The level of conﬁdence decreases

20 / 41 Tests of Signiﬁcance Why we need them (and some terminology)

Signiﬁcance tests are a formal way for us to draw conclusions about a statement using observed data. These tests are the tools we use to investigate whether the data corroborate the hypothesis that is being put forth. A hypothesis is a statement about the parameters in a population or model.

Null hypothesis, H0 The statement or hypothesis being tested in a signiﬁcance test is called the null hypothesis, and is normally referred to as H0. The signiﬁcance tests assess the evidence for and against this hypothesis and allow us to either reject or fail to reject H0.

Consider the following example to see how we can put these to use.

21 / 41 Tests of Signiﬁcance

Example: Are the bottles being ﬁlled as advertised? Suppose we are appointed as inspectors at an Irn Bru factory here in Scotland. We have data on past production and observe that the distribution of the contents is normal with standard deviation of 2ml. To assess the bottling process, we randomly select 10 bottles, measure their contents and obtain the following results:

502.9 499.8 503.2 502.8 500.9 503.9 498.2 502.5 503.8 501.4

Example: Are the bottles being ﬁlled as advertised? The randomly selected bottles resulted in the following measurements (and recall that σ = 2).

502.9 499.8 503.2 502.8 500.9 503.9 498.2 502.5 503.8 501.4 22 / 41 For this sample of observations, the mean content, x¯, is 501.94 ml. Is this sample mean far enough from 500 ml to provide convincing evidence that the mean content of all bottles produced at the factory differ from the advertised amount of 500 ml? Two incorrect ways of approaching this question is to say: 1. The mean of the sampled bottles is different from 500 ml, so the process is not filling the bottles at the correct mean level. 2. The difference of 1.94 ml is small relative to 500 ml, so this result is not surprising and the process is working well. Example: Are the bottles being filled as advertised? The randomly selected bottles resulted in the following measurements (and recall that σ = 2).

502.9 499.8 503.2 502.8 500.9 503.9 498.2 502.5 503.8 501.4

For this sample of observations, the mean content, x¯, is 501.94 ml. Is this sample mean far enough from 500 ml to provide convincing evidence that the mean content of all bottles produced at the factory differ from the advertised amount of 500 ml? Instead, we should formulate our null hypothesis and the "competing" or "alternative" hypothesis, and conduct a formal test of significance. In this example, H0 : µ = 500 and Ha : µ 6= 500 And all we are missing is a test statistic that will help us assess how much evidence there is towards the null hypothesis. Example: Are the bottles being filled as advertised? The randomly selected bottles resulted in the following measurements (and recall that σ = 2).

502.9 499.8 503.2 502.8 500.9 503.9 498.2 502.5 503.8 501.4

For this sample of observations, the mean content, x¯, is 501.94 ml. Is this sample mean far enough from 500 ml to provide convincing evidence that the mean content of all bottles produced at the factory diﬀer from the advertised amount of 500 ml? To obtain the test statistic, we simply use the standardised version of x¯: x¯ − µ 501.94 − 500 z = = = 3.07(> 1.96) √σ √2 n 10 We already know that 3.07 is quite far away from the mean of the standard normal distribution (which is, of course, 0). In fact, 95% of the area under the standard normal curve lies within [−1.96, 1.96]. Therefore, using the observed sample of 10 bottles, we reject H0 and conclude that our data are not compatible with H0. Example: Are the bottles being ﬁlled as advertised? The randomly selected bottles resulted in the following measurements (and recall that σ = 2).

502.9 499.8 503.2 502.8 500.9 503.9 498.2 502.5 503.8 501.4

As we mentioned earlier, test statistics provide us with a measure of evidence against the null hypothesis. The farther away the observations from what we would expect if the H0 were true, the more evidence there is against H0. The z-statistic we calculated earlier is one way of measuring how far the data are from what we would expect, which is what allows us to draw conclusions about the hypotheses. Another way to quantify how far away from what we expect the data are is using a p-value. P-value The p-value is the probability that the corresponding test statistic would take a value as or more extreme than that is actually observed assuming H0 is true. Hence, the smaller the p-value, the stronger the evidence in the data against H0 being true.

23 / 41 Tests of Signiﬁcance Calculating p-values Consider our earlier example about the Irn Bru factory, where we calculated the z-statistic to be 3.07 using our sample of size n = 10, standard deviation of 2 ml and the sample mean x¯ = 501.94: x¯ − µ 501.94 − 500 z = = = 3.07 √σ √2 n 10

If H0 is true, we expect z to be close to 0. If z is far from 0, there’s evidence against H0. Then, the p-value is P = Pr(z ≤ −3.07) + Pr(z ≥ 3.07). Since z has the standard normal distribution (N (0, 1)) under H0, we can find the area under the standard normal probability density function (pdf). NB: We only need to find one of the areas above because the standard normal pdf is symmetric around 0. 24 / 41 Tests of Significance Calculating p-values We can find this area from a standard normal table:

25 / 41 Tests of Signiﬁcance Calculating p-values

Pr(z ≤ −3.07) = 0.0011 (from table A) Pr(z ≥ 3.07) = 1 − Pr(z ≤ 3.07) = 1 − 0.9989 = 0.0011 = Pr(z ≤ −3.07) P = Pr(z ≤ −3.07) + Pr(z ≥ 3.07) = 2Pr(Z ≤ −3.07) = 0.0022

Now that we obtained the p-value, we need to decide what level of signiﬁcance to use in our test. The signiﬁcance level determines how much evidence we require to reject H0, and is usually denoted by the Greek letter alpha, α.

If we choose α = 0.05, rejecting H0 requires evidence so strong that it would happen no more than 5% of the time if H0 is true. If we choose α = 0.01, we require even stronger evidence against H0 to be able to reject it: evidence against H0 would need to be so strong that it would happen only 1% of the time if H0 is true.

26 / 41 Tests of Signiﬁcance P-values and statistical signiﬁcance

Statistical signiﬁcance If the p-value we calculate is smaller than our chosen α, we reject H0 at signiﬁcance level α.

Based on this, in our earlier example, we reject the null hypothesis, H0, at the 5% level because our p-value of 0.0022 < 0.05.

In fact, we reject H0 even at the 1% level because 0.0022 < 0.01. This suggests that there is very strong evidence against the null hypothesis, because if H0 were true, the observed sample mean should not have happened any more than 1% of the time.

27 / 41 Tests of Signiﬁcance One- and two-sided alternative hypotheses So far, we have focused on two-sided alternative hypotheses where Ha is stated as µ 6= c, where c is a constant. In some cases, however, we might be interested in a one-sided alternative hypothesis. In such cases, Ha would be expressed as µ < c or µ > c. In some cases, however, we might be interested in a one-sided alternative hypothesis. In such cases, Ha would be expressed as µ < c or µ > c. The p-values then look like this:

28 / 41 Tests of Signiﬁcance Summary 1/2

I Signiﬁcance tests allow us to formally assess the evidence against a null hypothesis (H0) provided by data. This way, we can judge whether the deviations from what the null hypothesis suggests are due to chance.

I When stating hypotheses, H0 is usually a statement that no effect exists (e.g. all bottles at a factory are filled with a mean quantity of 500 ml). The alternative hypothesis, Ha, on the other hand, suggests that a parameter differs from its null value in either direction (two-sided alternative) or in a specific direction (one-sided alternative).

I The test itself is conducted using a test statistic. The corresponding p-value is calculated assuming H0 is true, and it indicates the probability that the test statistic will take a value at least as "surprising" as the observed one.

29 / 41 Tests of Signiﬁcance Summary 2/2

I Small p-values indicate strong evidence against H0. I If the p-value is smaller than a specified significance level, α, we reject H0 and conclude that the data are statistically significant at level α.

I The test statistic concerning an unknown mean, µ, of a population are based on the z-statistic calculated as: x¯ − µ z = √σ n

where x¯ is the sample mean, n is the sample size, and σ is the known population standard deviation.

30 / 41 Tests of Signiﬁcance Standard error Consider a sample of size n from a normally distributed population with mean µ and standard deviation σ. As always, the sample mean is x¯ and is normally distributed with mean µ and standard √σ deviation n . When σ is unknown, we estimate it using the sample standard deviation, s. We can then estimate the standard deviation of x¯ as √s n . This quantity is called the standard error of the sample mean, x¯. Standard Error Standard deviation of a statistic estimated from the data is referred to as the standard error of the statistic. The standard error of the sample mean is s SE = √ n

31 / 41 Tests of Signiﬁcance z versus t distribution When σ is known, we can use the familiar z statistic to make x¯−µ inferences about µ: z = √σ . Recall that this standardized n statistic has the standard normal distribution, N (0, 1). When σ is not known, however, we need to estimate it using the sample standard deviation, s. As a result, we substitute the √σ √s standard deviation n with the standard error n and our test statistic becomes x¯ − µ t = √s n This statistic does not have a standard normal distribution, and instead follows what’s called a t distribution. There are lots of t distributions depending on its degrees of freedom.

32 / 41 Tests of Signiﬁcance t distributions

33 / 41 Tests of Signiﬁcance z versus t distribution To use the appropriate t distribution, we need to know the correct degrees of freedom. In the case of our simple t statistic, this is simply n − 1, where n is the number of observations in our data set. This is because the degrees of freedom come from the sample standard deviation, s, which has n − 1 degrees of freedom. Question: Why does s have n − 1 degrees of freedom? q 1 PN 2 To see why, note that s = N−1 i=1(xi − x¯) and that PN i=1(xi − x¯) = 0 (i.e. the deviations from the mean sum to zero). This implies that if we know n − 1 of the deviations, we can determine the last one. So, technically, only n − 1 deviations can vary freely and this number (i.e. n − 1) is called the degrees of freedom.

34 / 41 Conﬁdence Intervals ...using t distributions Using t distributions allows us to analyze samples from normally distributed populations without the need to know σ.

√σ As we saw earlier, replacing the standard deviation n of x¯ by its √s standard error n readily converts the z statistic into a t statistic. All hypothesis tests and conﬁdence intervals can be conducted/obtained the same way as before simply by using the appropriate t distribution (i.e. using the correct degrees of freedom).

√σ More speciﬁcally, the margin of error that was z × n becomes √s t × n .

35 / 41 Conﬁdence Intervals ...using t distributions And formally,

t Conﬁdence Interval A level C conﬁdence interval for a population mean (µ) is s x¯ ± t∗ × √ n where t∗ is the critical value with area C between −t∗ and t∗ under the t(n − 1) density curve, and n − 1 is the degrees of freedom.

36 / 41 Conﬁdence Intervals ...using t distributions

Example Here are monthly dollar amounts for phone service for a random sample of 8 households: 43, 47, 51, 36, 50, 42, 37, 41. We would like to construct a 95% CI for the average monthly expenditure, µ.

The sample mean is 43 + 47 + ··· + 41 x¯ = = 43.5 8 and the standard deviation is s (43 − 43.5)2 + (47 − 43.5)2 + ··· + (41 − 43.5)2 s = = 5.42 8 − 1 37 / 41 with degrees of freedom n − 1 = 7. Then, the standard error of x¯ is s 5.42 SE = √ = √ = 1.92 n 8 Conﬁdence Intervals ...using t distributions From the t distribution table, we ﬁnd that t∗ = 2.365, and thus the margin of error is

m = 2.365 × SE = (2.365)(1.92) = 4.5

And the 95% conﬁdence interval is

x¯ ± m = 43.5 ± 4.5 = (39, 48)

38 / 41 Conﬁdence Intervals ...using t distributions

39 / 41 Hypothesis tests ...using t distributions

Example Suppose that the overall U.S. average monthly expenditure for phone service is $49. Is the sample mean, x¯, of 43.5 diﬀerent from the national average of $49?

Before we attempt the question, we should state our hypotheses:

H0 : µ = 49

Ha : µ 6= 49 Then, note that x¯ = 43.5, n = 8, and s = 5.42. So, the t test statistic is

x¯ − µ0 43.5 − 49 t = s = = −2.87 √ 5√.42 n 8

Example Suppose that the overall U.S. average monthly expenditure for phone service is 40 / 41 $49. Is the sample mean, x¯, of 43.5 diﬀerent from the national average of $49?

Before we attempt the question, we should state our hypotheses:

H0 : µ = 49

Ha : µ 6= 49 Then, note that x¯ = 43.5, n = 8, and s = 5.42. So, the t test statistic is

x¯ − µ0 43.5 − 49 t = s = = −2.87 √ 5√.42 n 8 Because we have 7 degrees of freedom, this t statistic has the t(7) distribution. The p-value for this two-sided test is 2P(T ≥ 2.87). From the t table, we see that P(T ≥ 2.517) = 0.02 and P(T ≥ 2.998) = 0.01. Hence, our p-value is between 2 × 0.01 = 0.02 and 2 × 0.02 = 0.04. We can obtain the exact value using a computer, which yields P = 0.0241. This p-value suggests that the sample mean, x¯, of $43.5 is statistically signiﬁcantly diﬀerent from the nation-wide average of $49 at the 5% level, and that the randomly selected sample is not in line with the national average. References DeGroot, M. H. & Schervish, M. J. (2002). Probability and Statistics. Addison-Wesley. Moore, D. et al. (2011). The Practice of Statistics for Business and Economics. Third Edition. W. H. Freeman. Stock, J. H. & Watson, M. W. (2010). Introduction to Econometrics. Third Edition. Pearson Education.

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