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Using Permutations to Calculate Probabilities

By Jim Frost 6 Comments

Permutations in probability theory and other branches of mathematics refer to sequences of outcomes where the order matters. For example, 9-6-8-4 is a permutation of a four-digit PIN because the order of numbers is crucial. When calculating probabilities, it’s frequently necessary to calculate the number of possible permutations to determine an event’s probability.

In this post, I explain permutations and show how to calculate the number of permutations both with repetition and without repetition. Finally, we’ll work through a step-by-step example problem that uses permutations to calculate a probability. [Read more…] about Using Permutations to Calculate Probabilities

Filed Under: Probability Tagged With: analysis example, choosing analysis, conceptual

Understanding Historians’ Rankings of U.S. Presidents using Regression Models

By Jim Frost 6 Comments

Historians rank the U.S. Presidents from best to worse using all the historical knowledge at their disposal. Frequently, groups, such as C-Span, ask these historians to rank the Presidents and average the results together to help reduce bias. The idea is to produce a set of rankings that incorporates a broad range of historians, a vast array of information, and a historical perspective. These rankings include informed assessments of each President’s effectiveness, leadership, moral authority, administrative skills, economic management, vision, and so on. [Read more…] about Understanding Historians’ Rankings of U.S. Presidents using Regression Models

Filed Under: Regression Tagged With: analysis example, graphs, interpreting results

Spearman’s Correlation Explained

By Jim Frost 15 Comments

Spearman’s correlation in statistics is a nonparametric alternative to Pearson’s correlation. Use Spearman’s correlation for data that follow curvilinear, monotonic relationships and for ordinal data. Statisticians also refer to Spearman’s rank order correlation coefficient as Spearman’s ρ (rho).

In this post, I’ll cover what all that means so you know when and why you should use Spearman’s correlation instead of the more common Pearson’s correlation. [Read more…] about Spearman’s Correlation Explained

Filed Under: Basics Tagged With: analysis example, choosing analysis, conceptual, data types, Excel, graphs

Multiplication Rule for Calculating Probabilities

By Jim Frost 7 Comments

The multiplication rule in probability allows you to calculate the probability of multiple events occurring together using known probabilities of those events individually. There are two forms of this rule, the specific and general multiplication rules.

In this post, learn about when and how to use both the specific and general multiplication rules. Additionally, I’ll use and explain the standard notation for probabilities throughout, helping you learn how to interpret it. We’ll work through several example problems so you can see them in action. There’s even a bonus problem at the end! [Read more…] about Multiplication Rule for Calculating Probabilities

Filed Under: Probability Tagged With: analysis example, choosing analysis, conceptual

Exponential Smoothing for Time Series Forecasting

By Jim Frost 4 Comments

Exponential smoothing is a forecasting method for univariate time series data. This method produces forecasts that are weighted averages of past observations where the weights of older observations exponentially decrease. Forms of exponential smoothing extend the analysis to model data with trends and seasonal components. [Read more…] about Exponential Smoothing for Time Series Forecasting

Filed Under: Time Series Tagged With: analysis example, graphs, interpreting results

Descriptive Statistics in Excel

By Jim Frost 22 Comments

Descriptive statistics summarize your dataset, painting a picture of its properties. These properties include various central tendency and variability measures, distribution properties, outlier detection, and other information. Unlike inferential statistics, descriptive statistics only describe your dataset’s characteristics and do not attempt to generalize from a sample to a population. [Read more…] about Descriptive Statistics in Excel

Filed Under: Basics Tagged With: analysis example, Excel, interpreting results

Using Contingency Tables to Calculate Probabilities

By Jim Frost 13 Comments

Contingency tables are a great way to classify outcomes and calculate different types of probabilities. These tables contain rows and columns that display bivariate frequencies of categorical data. Analysts also refer to contingency tables as crosstabulation (cross tabs), two-way tables, and frequency tables.

Statisticians use contingency tables for a variety of reasons. I love these tables because they both organize your data and allow you to answer a diverse set of questions. In this post, I focus on using them to calculate different types of probabilities. These probabilities include joint, marginal, and conditional probabilities. [Read more…] about Using Contingency Tables to Calculate Probabilities

Filed Under: Probability Tagged With: analysis example, conceptual

How to Perform Regression Analysis using Excel

By Jim Frost 14 Comments

Excel can perform various statistical analyses, including regression analysis. It is a great option because nearly everyone can access Excel. This post is an excellent introduction to performing and interpreting regression analysis, even if Excel isn’t your primary statistical software package.

[Read more…] about How to Perform Regression Analysis using Excel

Filed Under: Regression Tagged With: analysis example, Excel, interpreting results

Independent and Dependent Samples in Statistics

By Jim Frost 12 Comments

When comparing groups in your data, you can have either independent or dependent samples. The type of samples in your design impacts sample size requirements, statistical power, the proper analysis, and even your study’s costs. Understanding the implications of each type of sample can help you design a better study. [Read more…] about Independent and Dependent Samples in Statistics

Filed Under: Basics Tagged With: analysis example, choosing analysis, conceptual

Using Moving Averages to Smooth Time Series Data

By Jim Frost 9 Comments

Moving averages can smooth time series data, reveal underlying trends, and identify components for use in statistical modeling. Smoothing is the process of removing random variations that appear as coarseness in a plot of raw time series data. It reduces the noise to emphasize the signal that can contain trends and cycles. Analysts also refer to the smoothing process as filtering the data. [Read more…] about Using Moving Averages to Smooth Time Series Data

Filed Under: Time Series Tagged With: analysis example, conceptual, Excel

A Tour of Survival Analysis

By Alexander Moreno 3 Comments

Note: this is a guest post by Alexander Moreno, a Computer Science PhD student at the Georgia Institute of Technology. He blogs at www.boostedml.com

Survival analysis is an important subfield of statistics and biostatistics. These methods involve modeling the time to a first event such as death. In this post we give a brief tour of survival analysis. We first describe the motivation for survival analysis, and then describe the hazard and survival functions. We follow this with non-parametric estimation via the Kaplan Meier estimator.  Then we describe Cox’s proportional hazard model and after that Aalen’s additive model. Finally, we conclude with a brief discussion.

Why Survival Analysis: Right Censoring

Modeling first event times is important in many applications. This could be time to death for severe health conditions or time to failure of a mechanical system. If one always observed the event time and it was guaranteed to occur, one could model the distribution directly. For instance, in the non-parametric setting, one could use the empirical cumulative distribution function to estimate the probability of death by some time. In the parametric setting one could do non-negative regression.

However, in some cases one might not observe the event time: this is generally called right censoring. In clinical trials with death as the event, this occurs when one of the following happens. 1) participants drop out of the study 2) the study reaches a pre-determined end time, and some participants have survived until the end 3) the study ends when a certain number of participants have died. In each case, after the surviving participants have left the study, we don’t know what happens to them. We then have the question:

  • How can we model the empirical distribution or do non-negative regression when for some individuals, we only observe a lower bound on their event time?

The above figure illustrates right censoring. For participant 1 we see when they died. Participant 2 dropped out, and we know that they survived until then, but don’t know what happened afterwards. For participant 3, we know that they survived until the pre-determined study end, but again don’t know what happened afterwards.

The Survival Function and the Hazard

Two of the key tools in survival analysis are the survival function and the hazard. The survival function describes the probability of the event not having happened by a time t. The hazard describes the instantaneous rate of the first event at any time t.

More formally, let T be the event time of interest, such as the death time. Then the survival function is S(t)=P(T>t). We can also note that this is related to the cumulative distribution function F(t)=P(T\leq t) via S(t)=1-F(t).

For the hazard, the probability of the first event time being in the small interval [t,t+dt), given survival up to t is P(T\in [t,t+dt)|T\geq t)=\lambda(t)dt. This is illustrated in the following figure.

Rearranging terms and taking limits we obtain

\lambda(t)=\lim_{dt\rightarrow 0}\frac{P(T\in [t,t+dt)|T\geq t)}{dt}=\frac{f(t)}{S(t)}

where f(t) is the density function of T and the second equality follows from applying Bayes theorem. By rearranging again and solving a differential equation, we can use the hazard to compute the survival function via

S(t)=\exp(-\int_0^t \lambda(s)ds)

The key question then is how to estimate the hazard and/or survival function.

Non-Parametric Estimation with Kaplan Meier

In non-parametric survival analysis, we want to estimate the survival function S(t) without covariates, and with censoring. If we didn’t have censoring, we could start with the empirical CDF \hat{F}(t)=\sum_{i=1}^n I(T_i\leq t). This equation is a succinct representation of: how many people have died by time t? The survival function would then be: how many people are still alive? However, we can’t answer this question as posed when some people are censored by time t.

While we don’t necessarily know how many people have survived by an arbitrary time t, we do know how many people in the study are still at risk. We can use this instead. Partition the study time into 0<t_1<\cdots t_{n-1}<t_n, where each t_i is either an event time or a censoring time for a participant. Assume that participants can only lapse at observed event times. Let Y(t) be the number of people at risk at just before time t. Assuming no one dies at exactly the same time (no ties), we can look at each time someone died. We say that the probability of dying at that specific time is \frac{1}{Y(t)}, and say that the probability of dying at any other time is 0. We can then say that the probability of surviving at any event time T_i, given survival at previous candidate event times is 1-\frac{1}{Y(T_i)}. The probability of surviving up to a time t is then

S(t)=\prod_{T_i\leq t}(1-\frac{1}{Y(T_i)})

We call this [1] the Kaplan Meier estimator. Under mild assumptions, including that participants have independent and identically distributed event times and that censoring and event times are independent, this gives an estimator that is consistent. The next figure gives an example of the Kaplan Meier estimator for a simple case.

Kaplan Meier R Example

In R we can use the Surv and survfit functions from the survival package to fit a Kaplan Meier model. We can also use ggsurvplot from the survminer package to make plots. Here we will use the ovarian cancer dataset from the survival package. We will stratify based on treatment group assignment.

library(survminer)
library(survival)
kaplan_meier <- Surv(time = ovarian[['futime']], event = ovarian[['fustat']])
kaplan_meier_treatment<-survfit(kaplan_meier~rx,data=ovarian, type='kaplan-meier',conf.type='log')
ggsurvplot(kaplan_meier_treatment,conf.int = 'True')

Semi-Parametric Regression with Cox’s Proportional Hazards Model

Kaplan Meier makes sense when we don’t have covariates, but often we want to model how some covariates affect death risk. For instance, how does one’s weight affect death risk? One way to do this is to assume that covariates have a multiplicative effect on the hazard. This leads us to Cox’s proportional hazard model, which involves the following functional form for the hazard:

\lambda(t)=\lambda_0(t)\exp(\beta^T x)

The baseline hazard \lambda_0(t) describes how the average person’s risk evolves over time. The relative risk \exp(\beta^T x) describes how covariates affect the hazard. In particular, a unit increase in x_i leads to an increase of the hazard by a factor of \exp(\beta_i).

Because of the non-parametric nuissance term \lambda_0(t), it is difficult to maximize the full likelihood for \beta directly. Cox’s insight [2] was that the assignment probabilities given the death times contain most of the information about \beta, and the remaining terms contain most of the information about \lambda_0(t). The assignment probabilities give the following partial likelihood

L(\beta)=\prod_{i=1}^n\frac{\exp(\beta^T x^{(i)})}{\sum_{k=1}^n \exp(\beta^T x^{(k)})}

We can then maximize this to get an estimator \hat{\beta} of \beta. In [3,4] they show that this estimator is consistent and asymptotically normal.

Cox Proportional Hazards R Example

In R, we can use the Surv and coxph functions from the survival package. For the ovarian cancer dataset, we notice from the Kaplan Meier example that treatment is not proportional. Under a proportional hazards assumption, the curves would have the same pattern but diverge. However, instead they move apart and then move back together. Further, treatment does seem to lead to different survival patterns over shorter time horizons. We should not use it as a covariate, but we can stratify based on it. In R we can regress on age and presence of residual disease.

cox_fit <- coxph(Surv(futime, fustat) ~ age + ecog.ps+strata(rx), data=ovarian)
summary(cox_fit)

which gives the following results

Call:
coxph(formula = Surv(futime, fustat) ~ age + ecog.ps + strata(rx), 
    data = ovarian)

  n= 26, number of events= 12 

            coef exp(coef) se(coef)      z Pr(>|z|)   
age      0.13853   1.14858  0.04801  2.885  0.00391 **
ecog.ps -0.09670   0.90783  0.62994 -0.154  0.87800   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

        exp(coef) exp(-coef) lower .95 upper .95
age        1.1486     0.8706    1.0454     1.262
ecog.ps    0.9078     1.1015    0.2641     3.120

Concordance= 0.819  (se = 0.058 )
Likelihood ratio test= 12.71  on 2 df,   p=0.002
Wald test            = 8.43  on 2 df,   p=0.01
Score (logrank) test = 12.24  on 2 df,   p=0.002


this suggests that age has a significant multiplicative effect on death, and that a one year increase in age increases instantaneous risk by a factor of 1.15.

Aalen’s Additive Model

Cox regression makes two strong assumptions: 1) that covariate effects are constant over time 2) that effects are multiplicative. Aalen’s additive model [5] relaxes the first, and replaces the second with the assumption that effects are additive. Here the hazard takes the form

\lambda_i(t)=\beta_0(t)+\beta_1(t)x^{(i)}_1+\cdots+\beta_p(t)x^{(i)}_p

As this is a linear model, we can estimate the cumulative regression functions using a least squares type procedure.

Aalen’s Additive Model R Example

In R we can use the timereg package and the aalen function to estimate cumulative regression functions, which we can also plot.

library(timereg)
data(sTRACE) 
# Fits Aalen model 
out<-aalen(Surv(time,status==9)~age+sex+diabetes+chf+vf, sTRACE,max.time=7,n.sim=100) 
summary(out) 
par(mfrow=c(2,3)) 
plot(out)

This gives us

Additive Aalen Model 

Test for nonparametric terms 

Test for non-significant effects 
            Supremum-test of significance p-value H_0: B(t)=0
(Intercept)                          7.29                0.00
age                                  8.63                0.00
sex                                  2.95                0.01
diabetes                             2.31                0.24
chf                                  5.30                0.00
vf                                   2.95                0.03

Test for time invariant effects 
                  Kolmogorov-Smirnov test
(Intercept)                       0.57700
age                               0.00866
sex                               0.11900
diabetes                          0.16200
chf                               0.12900
vf                                0.43500
            p-value H_0:constant effect
(Intercept)                        0.00
age                                0.00
sex                                0.18
diabetes                           0.43
chf                                0.06
vf                                 0.02
                    Cramer von Mises test
(Intercept)                      0.875000
age                              0.000179
sex                              0.017700
diabetes                         0.041200
chf                              0.053500
vf                               0.434000
            p-value H_0:constant effect
(Intercept)                        0.00
age                                0.00
sex                                0.29
diabetes                           0.42
chf                                0.02
vf                                 0.05

   
   
  Call: 
aalen(formula = Surv(time, status == 9) ~ age + sex + diabetes + 
    chf + vf, data = sTRACE, max.time = 7, n.sim = 100)

The results first test whether the cumulative regression functions are non-zero, and then whether the effects are constant. The plots of the cumulative regression functions are given below.

Discussion

In this post we did a brief tour of several methods in survival analysis. We first described why right censoring requires us to develop new tools. We then described the survival function and the hazard. Next we discussed the non-parametric Kaplan Meier estimator and the semi-parametric Cox regression model. We concluded with Aalen’s additive model.

[1] Kaplan, Edward L., and Paul Meier. “Nonparametric estimation from incomplete observations.” Journal of the American statistical association 53, no. 282 (1958): 457-481.
[2] Cox, David R. “Regression models and life-tables.” In Breakthroughs in statistics, pp. 527-541. Springer, New York, NY, 1992.
[3] Tsiatis, Anastasios A. “A large sample study of Cox’s regression model.” The Annals of Statistics 9, no. 1 (1981): 93-108.
[4] Andersen, Per Kragh, and Richard David Gill. “Cox’s regression model for counting processes: a large sample study.” The annals of statistics (1982): 1100-1120.
[5] Aalen, Odd. “A model for nonparametric regression analysis of counting processes.” In Mathematical statistics and probability theory, pp. 1-25. Springer, New York, NY, 1980.

Filed Under: Survival Tagged With: analysis example, choosing analysis, conceptual

How the Chi-Squared Test of Independence Works

By Jim Frost 17 Comments

Chi-squared tests of independence determine whether a relationship exists between two categorical variables. Do the values of one categorical variable depend on the value of the other categorical variable? If the two variables are independent, knowing the value of one variable provides no information about the value of the other variable.

I’ve previously written about Pearson’s chi-square test of independence using a fun Star Trek example. Are the uniform colors related to the chances of dying? You can test the notion that the infamous red shirts have a higher likelihood of dying. In that post, I focus on the purpose of the test, applied it to this example, and interpreted the results.

In this post, I’ll take a bit of a different approach. I’ll show you the nuts and bolts of how to calculate the expected values, chi-square value, and degrees of freedom. Then you’ll learn how to use the chi-squared distribution in conjunction with the degrees of freedom to calculate the p-value. [Read more…] about How the Chi-Squared Test of Independence Works

Filed Under: Hypothesis Testing Tagged With: analysis example, distributions, interpreting results

How to Test Variances in Excel

By Jim Frost 6 Comments

Use a variances test to determine whether the variability of two groups differs. In this post, we’ll work through a two-sample variances test that Excel provides. Even if Excel isn’t your primary statistical software, this post provides an excellent introduction to variance tests. Excel refers to this analysis as F-Test Two-Sample for Variances. [Read more…] about How to Test Variances in Excel

Filed Under: Hypothesis Testing Tagged With: analysis example, Excel, interpreting results

How to do Two-Way ANOVA in Excel

By Jim Frost 22 Comments

Use two-way ANOVA to assess differences between the group means that are defined by two categorical factors. In this post, we’ll work through two-way ANOVA using Excel. Even if Excel isn’t your main statistical package, this post is an excellent introduction to two-way ANOVA. Excel refers to this analysis as two factor ANOVA. [Read more…] about How to do Two-Way ANOVA in Excel

Filed Under: ANOVA Tagged With: analysis example, Excel, interpreting results

5 Ways to Find Outliers in Your Data

By Jim Frost 24 Comments

Outliers are data points that are far from other data points. In other words, they’re unusual values in a dataset. Outliers are problematic for many statistical analyses because they can cause tests to either miss significant findings or distort real results.

Unfortunately, there are no strict statistical rules for definitively identifying outliers. Finding outliers depends on subject-area knowledge and an understanding of the data collection process. While there is no solid mathematical definition, there are guidelines and statistical tests you can use to find outlier candidates. [Read more…] about 5 Ways to Find Outliers in Your Data

Filed Under: Basics Tagged With: analysis example, conceptual, graphs

How to do One-Way ANOVA in Excel

By Jim Frost 21 Comments

Use one-way ANOVA to determine whether the means of at least three groups are different. Excel refers to this test as Single Factor ANOVA. This post is an excellent introduction to performing and interpreting one-way ANOVA even if Excel isn’t your primary statistical software package. [Read more…] about How to do One-Way ANOVA in Excel

Filed Under: ANOVA Tagged With: analysis example, Excel, interpreting results

How to do t-Tests in Excel

By Jim Frost 79 Comments

Excel can perform various statistical analyses, including t-tests. It is an excellent option because nearly everyone can access Excel. This post is a great introduction to performing and interpreting t-tests even if Excel isn’t your primary statistical software package.

In this post, I provide step-by-step instructions for using Excel to perform t-tests. Importantly, I also show you how to select the correct form of t-test, choose the right options, and interpret the results. I also include links to additional resources I’ve written, which present clear explanations of relevant t-test concepts that you won’t find in Excel’s documentation. And, I use an example dataset for us to work through and interpret together! [Read more…] about How to do t-Tests in Excel

Filed Under: Hypothesis Testing Tagged With: analysis example, Excel, interpreting results

Revisiting the Monty Hall Problem with Hypothesis Testing

By Jim Frost 9 Comments

The Monty Hall Problem is where Monty presents you with three doors, one of which contains a prize. He asks you to pick one door, which remains closed. Monty opens one of the other doors that does not have the prize. This process leaves two unopened doors—your original choice and one other. He allows you to switch from your initial choice to the other unopened door. Do you accept the offer?

If you accept his offer to switch doors, you’re twice as likely to win—66% versus 33%—than if you stay with your original choice.

Mind-blowing, right?

The solution to the Monty Hall Problem is tricky and counter-intuitive. It did trip up many experts back in the 1980s. However, the correct answer to the Monty Hall Problem is now well established using a variety of methods. It has been proven mathematically, with computer simulations, and empirical experiments, including on television by both the Mythbusters (CONFIRMED!) and James Mays’ Man Lab. You won’t find any statisticians who disagree with the solution.

In this post, I’ll explore aspects of this problem that have arisen in discussions with some stubborn resisters to the notion that you can increase your chances of winning by switching!

The Monty Hall problem provides a fun way to explore issues that relate to hypothesis testing. I’ve got a lot of fun lined up for this post, including the following!

  • Using a computer simulation to play the game 10,000 times.
  • Assessing sampling distributions to compare the 66% percent hypothesis to another contender.
  • Performing a power and sample size analysis to determine the number of times you need to play the Monty Hall game to get an answer.
  • Conducting an experiment by playing the game repeatedly myself, record the results, and use a proportions hypothesis test to draw conclusions! [Read more…] about Revisiting the Monty Hall Problem with Hypothesis Testing

Filed Under: Hypothesis Testing Tagged With: analysis example, conceptual, distributions, interpreting results

Using Post Hoc Tests with ANOVA

By Jim Frost 76 Comments

Post hoc tests are an integral part of ANOVA. When you use ANOVA to test the equality of at least three group means, statistically significant results indicate that not all of the group means are equal. However, ANOVA results do not identify which particular differences between pairs of means are significant. Use post hoc tests to explore differences between multiple group means while controlling the experiment-wise error rate.

In this post, I’ll show you what post hoc analyses are, the critical benefits they provide, and help you choose the correct one for your study. Additionally, I’ll show why failure to control the experiment-wise error rate will cause you to have severe doubts about your results. [Read more…] about Using Post Hoc Tests with ANOVA

Filed Under: ANOVA Tagged With: analysis example, choosing analysis, conceptual, graphs, interpreting results

One-Tailed and Two-Tailed Hypothesis Tests Explained

By Jim Frost 52 Comments

Choosing whether to perform a one-tailed or a two-tailed hypothesis test is one of the methodology decisions you might need to make for your statistical analysis. This choice can have critical implications for the types of effects it can detect, the statistical power of the test, and potential errors.

In this post, you’ll learn about the differences between one-tailed and two-tailed hypothesis tests and their advantages and disadvantages. I include examples of both types of statistical tests. In my next post, I cover the decision between one and two-tailed tests in more detail.
[Read more…] about One-Tailed and Two-Tailed Hypothesis Tests Explained

Filed Under: Hypothesis Testing Tagged With: analysis example, conceptual, interpreting results

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