Lecture 4
Administrative Miscellanea
- Homework 4 due Friday
- Quiz 2 today
- Problem Set 1 Graded - Partial Credit available until Friday
- Exam next Wednesday
- Study materials will be available this week
Hypothesis testing: motivating question
- Run \(y_i=\beta_0+\beta_1 x_i + \varepsilon_i\)
- Suppose \(\beta_1=0\) (the true value), will we obtain \(\hat\beta_1=0\)?
- No. Even without bias we will still have some random variation due to sampling
- We know that \(\hat\beta_1\sim N(\beta_1,\frac{\hat\sigma_\varepsilon}{n\sigma_x^2})\)
Probabilities under continuous random variables
- This is a probability density function
- What is the probability that X=0?
- We’ll see why in a second
Probabilities under continuous random variables
- We want to find the area of the shaded region (between -1 and 1)
- The function pnorm \((=\Phi(z))\) gives \(P(X\le z)\)
- How do we calculate \(P(-1\le x\le 1)?\)
Probabilities under continuous random variables
- We grab the full area to the left of 1 (pnrom(1) = green + red area), then subtract out the area to the left of -1 (pnorm(-1))
Question: Probabilities
- \(\hat\beta_1\sim N(0,1)\). What is the probability that \(\hat\beta_1\) is between -2 and 1, \(P(-2<\hat\beta_1<1)\)?
x area_to_left
1: -3.0 0.001
2: -2.5 0.006
3: -2.0 0.023
4: -1.5 0.067
5: -1.0 0.159
6: -0.5 0.309
7: 0.0 0.500
8: 0.5 0.691
9: 1.0 0.841
10: 1.5 0.933
11: 2.0 0.977
12: 2.5 0.994
13: 3.0 0.999Answer: Probabilities
Question: calculating z scores
- Given a standard normal (\(Z \sim N(0,1)\)) we can calculate \(P(a<X<b)\) with \(\Phi(b)-\Phi(a)\) where \(\Phi(x)\) is the pnorm function we used earlier
- If \(X\sim N(\mu,\sigma)\) how we calculate \(P(a<X<b)?\)
- Standardize: subtract the mean and divide by the standard deviation to obtain the z score. This is still a normal random variable
- \(P(a<x<b) = P(\frac{a-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{b-\mu}{\sigma}) = P(z_a<Z<z_b)\)
- \(= \Phi(z_a)-\Phi(z_b)\)
Question: z scores normal
- what is the probability that \(\hat\beta_1\) is between -2 and 2, but now \(\hat\beta_1\sim N(1,2)\), ie it now has mean 1 and standard error of 2? First step?
- \((\hat\beta_1 -1)/2 \equiv z\sim N(0,1)\)
- Now translate the probability: \(-2<\beta_2<2\implies -1.5<z<.5\)
Answer: z scores normal
x area_to_left
1: -3.0 0.001
2: -2.5 0.006
3: -2.0 0.023
4: -1.5 0.067
5: -1.0 0.159
6: -0.5 0.309
7: 0.0 0.500
8: 0.5 0.691
9: 1.0 0.841
10: 1.5 0.933
11: 2.0 0.977
12: 2.5 0.994
13: 3.0 0.999[1] 0.6246553t distribution vs Z
- Asymptotically (as \(n\to\infty\)) \(\hat\beta_1\) is normally distributed, but for small n it actually follows a t distribution
- The t distribution has a mean and standard deviation, but also degrees of freedom (\(df=n-1\))
- Nothing actually changes, except when we lookup a table or calculate in R we use slightly different function to calculate
- In R: pt(z,df) instead of pnorm(z)
- It is rare to have small samples in econometrics - they basically give the same result
- R regression output gives you t values by default
T distribution: graph
Flipping the question
- We know that if \(\hat\beta_1=0\) we may still get nonzero results
- If we obtain \(\hat\beta_1=1\), what is the probability that \(\beta_1=0\)? i.e. what is the probability there is actually no effect but we obtain a nonzero result?
- This is impossible to answer, but we can get a suggestive answer from the prior exercise
- We assume the null hypothesis (\(\beta_1=0\)), and calculate the probability that we obtain a value of \(\hat\beta_1\ge 1\), ie \(\hat\beta_1\sim N(0,1), p=P(\hat\beta_1>1)=1-\Phi(1)\approx .32\),
Adding Hypothesis Testing Jargon
- We are testing the hypothesis that x causes y, ie that \(\beta_1\neq0\) (we are explicitly assuming exogeneity at the moment). We label these hypotheses.
- \(H_0\) is always the null hypothesis: that no effect exists. Here \(H_0: \beta_1=0\)
- \(H_1\) is the alternative hypothsis here that there is an effect: \(H_1: \beta_1\neq0\). We could have also explicitly tested \(\beta_1>0\) or \(\beta_1<0\)
Adding Hypothesis Testing Jargon
- We don’t have enough information to calculate probabilities. Instead we ask: if \(H_0\) is true, what is the probability that we observe a result as extreme as \(\hat\beta_1\)? Suppose we obtained an estimate of \(\hat\beta_1=1\)
- \(P(|\hat\beta_1|>1\ |\ \beta_1=0)\)
- This is called the p value, here \(p=1-.68=.32\)
Adding Hypothesis Testing Jargon
- We set up a criterion for rejecting \(H_0\) which we call the signficance level \(\alpha\) (normally .05 or .01). If the probability of obtaining such a result under the null is small, we reject the null and assume the effect is not statistical noise, otherwise we conservatively fail to reject \(H_0\)
- Here \(\alpha=.05 > p=.32\) so we fail to reject \(H_0\)
- There could still be an effect (on average it’s 1), it’s just too small to confidently conclude it wasn’t random chance
Adding Hypothesis Testing Jargon
- Under this setup, we will reject \(\alpha=5\%\) of cases where \(\beta_1=0\) by chance (the type I error rate, or false positive rate)
- We can also fail to reject \(H_0\) (and conclude there is no effect) even when \(\beta_1\neq 0\), called a type II error or false negative
- This occurs with probability \(\beta\), which requires additional assumptions to calculate
Hypothesis testing: Type 1 and Type 2 errors
- We can either reject or fail to reject \(H_0\), additional \(H_0\) can either be true or false (note that we can never observe this). This leads to 4 possibilities
Hypothesis testing: Type 1 and Type 2 errors
- \(H_0\) is true and you fail to reject H_0 : correct decision
- \(H_0\) is true and you reject H_0: False positive (type 1 error) This happens with probability \(\alpha\)
- \(H_0\) is false and you fail to reject \(H_0\): False negative (type 2 error). This happens with probability \(\beta\), which requires additional assumptions to calculate
- \(H_0\) is false and you reject \(H_0\) Correct decision
Power Curve
- Let \(\alpha=.05\), \(H_0: \hat\beta_1\sim N(0,1)\). Then we fail to reject \(H_0\) if \(-1.96<\hat\beta_1<1.96\)
- Probability of a type I error is .05 Type 2 error depends on what \(\beta_1 is\)
- Calculate this for each value of \(\beta_1\)
- Power curve graphs the “power” - \(1-\beta_1\) (so that a power of 1 corresponds to a type II error rate of 0)
- Because this is symmetric we normally only graph the right hand side (which we can reference as \(|\beta_1|\))
Power Curve: Graph
Power: Implications
- Suppose my outcome variable is number of years of education achieved. A result of .0001 years may end up being statistically significant, but not practically
- Suppose I determine that anything below 1 year is not very significant. Then I can look up \(\beta=1\) in the power curve and determine my probability of detecting an effect of that size
- In the prior graph I only have a 20% chance of being able to detect such a change. I need an effect of around 3 just to have an 80% chance of detecting an event
Power: Implications
- Given power, we can now interpret results of hypothesis testing. Assuming all of our modeling assumptions hold, then:
- If we reject \(H_0\) then we know this is likely a real effect, though we must interpret it since it could be incredibly small in practice
- If we fail to reject \(H_0\) then it may have just been random noise, but there also could have been a sizeable effect that was missed due to lower power (e.g. \(\beta=1\))
Power: Implications
- If we have a large sample size then our interpretation is usually clear: we know whether there is an effect, and we can judge the size based on practical significance
- This is conditional on all other assumptions! This is why economic papers focus most of their effort on methodology and very little on p values
Hypothesis Testing: Example
- We randomly assign students to classroom with either 10 or 20 students in an experiment to determine the effect of class size on test scores. We run the regression \(score_i=\beta_0+\beta_1 size_i + \varepsilon_i\). After running an OLS regression we obtain \(\hat\beta_1 = 0.05, se(\hat\beta_1)= 0.04\) We wish to know whether class size affects student test scores, and use a significance level of \(\alpha=.05\)
- Write \(H_0, H_1\) using a two tailed test
- Calculate the standardized values (z scores) for the area you’re testing
Hypothesis Testing: Example
- We randomly assign students to classroom with either 10 or 20 students in an experiment to determine the effect of class size on test scores. We run the regression \(score_i=\beta_0+\beta_1 size_i + \varepsilon_i\). After running an OLS regression we obtain \(\hat\beta_1 = 0.05, se(\hat\beta_1)= 0.04\) We wish to know whether class size affects student test scores, and use a significance level of \(\alpha=.05\)
- Compute the p value
- Determine your decision
Hypothesis Testing: Example
- \(H_0:\beta_1=0, H_1: \beta_1\neq 0\)
- \(z=.05/.04=1.25 \implies |z|<1.25\)
- \(p=1-(\Phi(1.25)-\Phi(-1.25)) = 1- .789=.211\)
- Fail to reject \(H_0\)
Errors: Examples; categorization
- You are given the following output from R on a regression of y vs x. Do you reject \(H_0: \beta_1=0\) at \(\alpha=.05\)?
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-2.21741 -0.65004 -0.04366 0.47332 3.08603
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.18679 0.09869 -1.893 0.0613 .
x 0.01065 0.10383 0.103 0.9185
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.986 on 98 degrees of freedom
Multiple R-squared: 0.0001074, Adjusted R-squared: -0.0101
F-statistic: 0.01053 on 1 and 98 DF, p-value: 0.9185Confidence Intervals
- Suppose we reject \(H_0\) and have \(\hat\beta_1=1, se(\hat\beta_1)=0.1\).
- We know that there is likely an effect (if all assumptions are met), and that the most likely value of \(\beta_1\) is 1
- We would like to know the likely range of values \(\beta_1\) could actually be
- Instead of testing a hypothesis, we can form an interval around \(\hat\beta_1\) that captures 95% (or 99%, etc) of the likely range of values it can take on.
Confidence Intervals
- The calculation is essentially the same, and we call it a confidence interval: \(CI_{0.95} = \hat\beta_1 \pm z_{.95}se(\hat\beta_1)\)
- \(-1.96<z_{.95}<1.96\) gives an area of \(95\%\), so we have \(\hat\beta_1=1\pm .196 = (.804,1.196)\)
- How do we obtain 1.96? qnorm(.975). \(.975=1-.05/2\), ie we take \(\alpha\), divide between our two tails, and lookup the value in our table
Confidence Intervals
- We say that we are 95% confident that the true value of \(\beta_1\) is between .804 and 1.196
- We don’t say that there’s a 95% chance that \(\beta_1\) is in this range because it isn’t true. We’re making a lot of assumptions when calculating this value
Confidence Intervals: Example
- \(\hat\beta_1=3,se(\hat\beta_1)=2\) calculate a 99% confidence interval of \(\hat\beta_1\)
- \(3\pm 2.576*2=3\pm 5.152 = (-2.152,8.152)\)
When Hypothesis testing goes wrong
- The tea example is quite compelling, but in the real world hypothesis testing is highly misinterpreted
- Some journals are now banning significance based hypothesis testing, and the American Statistical Association has had to put out statements regarding the misuse of p values
- When we run a regression model, our hypothesis that we are rejecting is not that \(\beta_1=0\) (if we reject that then it must be that \(\beta_1\neq0\) which implies there is a causal relationship). Rather, we are rejecting the claim that \(\beta_1=0\) and all of our additional modeling assumptions are correct
When Hypothesis testing goes wrong
- In particular, we assumed in calculating the distribution of \(\hat\beta_1\) that this was an unbiased estimate (exogeneity), and we calculated the variance using a formula that assumed uncorrelated error terms.
- When we reject \(H_0\) we are just saying that at least one one of these assumptions is (statistically) incorrect
- Note that if we have an extremely large sample size even slight differences will be significant (e.g. \(\beta_1=.001\neq0\)), but this means even very small modeling assumption errors will also result in significance. On big data you will almost always get p<.01, but that means virtually nothing
Cluster correlation
When Hypothesis testing goes wrong: real data
Call:
lm(formula = grade ~ attendance, data = dt)
Residuals:
Min 1Q Median 3Q Max
-63.514 -8.129 0.613 9.083 40.188
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 54.011 3.703 14.588 < 2e-16 ***
attendance 38.164 4.377 8.719 9.41e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 15.85 on 206 degrees of freedom
(1 observation deleted due to missingness)
Multiple R-squared: 0.2695, Adjusted R-squared: 0.266
F-statistic: 76.02 on 1 and 206 DF, p-value: 9.408e-16
When Hypothesis testing goes wrong: real data
- Our p value is 0.000000000000000941. Highly Significant!
- We reject \(H_0\) that \(\beta_1=0\), but we absolutely cannot claim that this is a causal effect because there is no exogeneity.
- Rather, we’re concluding that all of our assumptions are implausible. i.e. it almost certainly cannot be the case that \(\beta_1=0\) AND we have exogeneity in our model AND our error terms are uncorrelated AND our true model is linear.
- Note that this says almost nothing!
Yet another type of wrong
So is there any use for hypothesis testing
- First, you must show that your modeling assumptions are valid. For example, claim exogeneity by using a randomized control trial or other compelling quasi-experimental design
- Second, make conservative estimates in your other assumptions. e.g. overstating the variance in your model will be more convincing than understating your variance
- Third, interpret your results within the context of how precise your estimate is
- Significance testing is a small but important part of research
Hypothesis testing: Summary of steps
- Calculate \(\hat\beta_1\) and \(se(\hat\beta_1) (=\sigma_{\hat\beta_1})\) (this will be given to you as the output to a regression)
- Compute the z-score (the number of standard deviations from the mean) : \(z = \frac{\hat\beta_1-0}{se(\hat\beta_1)}\)
- Calculate \(P(|Z|\ge z)\), where \(Z\sim N(0,1)\)
- Look it up in a normal table, or use 1-(pnorm(z)-pnorm(-z)) in R
- Compare p to \(\alpha\). If \(p<\alpha\), reject \(H_0\), otherwise fail to reject \(H_0\)
Hypothesis testing: Summary of steps
- For a confidence interval, instead calculate \(\hat\beta_1 \pm z_\alpha*se(\hat\beta_1)\)
- \(z_\alpha\) is looked up via a table, or qnorm in R. It’s 1.96 for 95% confidence