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    \title{Testing Stroop Effect}
    
    
    

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    \section{Testing a perpetual
phenomenon}\label{testing-a-perpetual-phenomenon}

Author: Vivek V Dwivedi Dated: July 11, 2018

\subsection{On your mark ...}\label{on-your-mark-...}

In a Stroop task, participants are presented with a list of words, with
each word displayed in a color of ink. The participant's task is to say
out loud the color of the ink in which the word is printed. The task has
two conditions: a congruent words condition, and an incongruent words
condition. In the congruent words condition, the words being displayed
are color words whose names match the colors in which they are printed:
for example {RED}, {BLUE}. In the incongruent words condition, the words
displayed are color words whose names do not match the colors in which
they are printed: for example {PURPLE}, {ORANGE}. In each case, we
measure the time it takes to name the ink colors in equally-sized lists.
Each participant will go through and record a time from each condition.

\subsection{Get Set ...}\label{get-set-...}

We are looking at the response time for saying congruent or incongruent
words. Hence we can state that:

\begin{itemize}
\item
  Congruent and Incongruent words, or the type of word is our
  \textbf{independent variable}.
\item
  Time taken to say the color of ink in two lists of same size would be
  our \textbf{dependent variable}.
\end{itemize}

With our dependent and independent variables identified, let's set our
null and alternate hypothesis. A null hypothesis in this case would be
that there is no significant difference between the mean time taken to
say colors for congruent (μc) and incongruent (μi) words. Even though it
seems intutive that incongruent words will take longer time, my
alternate hypothesis is that there is a significant difference in time
taken for congruent and incongruent lists.

H0: μc = μi

HA: μc != μi

With these null and alternate hypothesis set, what kind of statistical
tests am I going to do?

\emph{Two tailed dependent samples t-test}

Since I am not concerened about the direction, I will certainly go for a
two tailed test. My dataset for this test contains 25 records and
parameters are not available to me. Given that t distribution is used
for small sample sizes with unknown population variance, I will go for a
dependent samples t-test and see if we can retain or reject the null
hypothesis. Also the same person is taking the test twice, so this falls
under the \emph{within-subject} or \emph{repeated-measures} tests.

\subsection{Go ...}\label{go-...}

Let's figure out some descriptive statistics about our dataset.

\begin{itemize}
\tightlist
\item
  Degrees of freedom (n -1): 24 - 1 = 23
\end{itemize}

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        \PY{k}{def} \PY{n+nf}{mean}\PY{p}{(}\PY{n}{data}\PY{p}{)}\PY{p}{:}
            \PY{k}{return} \PY{n+nb}{sum}\PY{p}{(}\PY{n}{data}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{:}\PY{n+nb}{len}\PY{p}{(}\PY{n}{data}\PY{p}{)}\PY{p}{]}\PY{p}{)}\PY{o}{/}\PY{n+nb}{len}\PY{p}{(}\PY{n}{data}\PY{p}{)}
            
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}3}]:} \PY{c+c1}{\PYZsh{} takes a list of numbers and returns the median}
        \PY{k}{def} \PY{n+nf}{median}\PY{p}{(}\PY{n}{data}\PY{p}{)}\PY{p}{:}
            \PY{n}{data}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{p}{)}
            \PY{n}{length} \PY{o}{=} \PY{n+nb}{len}\PY{p}{(}\PY{n}{data}\PY{p}{)}
        
            \PY{k}{if} \PY{n}{length} \PY{o}{\PYZpc{}} \PY{l+m+mi}{2} \PY{o}{==} \PY{l+m+mi}{0}\PY{p}{:}
                \PY{n}{median} \PY{o}{=} \PY{p}{(}\PY{p}{(}\PY{n}{data}\PY{p}{[}\PY{p}{(}\PY{n}{length}\PY{o}{/}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{]} \PY{o}{+} \PY{n}{data}\PY{p}{[}\PY{p}{(}\PY{n}{length}\PY{o}{/}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{1}\PY{p}{]}\PY{p}{)} \PY{o}{/} \PY{l+m+mi}{2}\PY{p}{)}
            \PY{k}{else}\PY{p}{:}
                \PY{n}{median} \PY{o}{=} \PY{n}{data}\PY{p}{[}\PY{p}{(}\PY{n}{length}\PY{o}{/}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{]}
        
            \PY{k}{return} \PY{n}{median}
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}4}]:} \PY{c+c1}{\PYZsh{} given a sample and sample mean, it returns the variance}
        \PY{c+c1}{\PYZsh{} for this very specific case, I am providing the mean as well, else it can be calculated}
        \PY{k}{def} \PY{n+nf}{variance}\PY{p}{(}\PY{n}{data}\PY{p}{,} \PY{n}{mean}\PY{p}{)}\PY{p}{:}
            \PY{c+c1}{\PYZsh{} squared difference from mean}
            \PY{n}{squared\PYZus{}difference} \PY{o}{=} \PY{p}{[}\PY{p}{(}\PY{n}{num} \PY{o}{\PYZhy{}} \PY{n}{mean}\PY{p}{)}\PY{o}{*}\PY{o}{*}\PY{l+m+mi}{2} \PY{k}{for} \PY{n}{num} \PY{o+ow}{in} \PY{n}{data}\PY{p}{]} 
            \PY{c+c1}{\PYZsh{} sum of squared difference by degrees of freedom or sample size \PYZhy{} 1}
            \PY{k}{return} \PY{n+nb}{sum}\PY{p}{(}\PY{n}{squared\PYZus{}difference}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{:}\PY{n+nb}{len}\PY{p}{(}\PY{n}{squared\PYZus{}difference}\PY{p}{)}\PY{p}{]}\PY{p}{)}\PY{o}{/}\PY{p}{(}\PY{n+nb}{len}\PY{p}{(}\PY{n}{squared\PYZus{}difference}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{1}\PY{p}{)}
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}5}]:} \PY{k}{def} \PY{n+nf}{describe\PYZus{}data}\PY{p}{(}\PY{n}{data}\PY{p}{,} \PY{n}{entity\PYZus{}type}\PY{p}{)}\PY{p}{:}
            \PY{n}{mean\PYZus{}calculated} \PY{o}{=} \PY{n}{mean}\PY{p}{(}\PY{n}{data}\PY{p}{)}
            \PY{n}{variance\PYZus{}calculated} \PY{o}{=} \PY{n}{variance}\PY{p}{(}\PY{n}{data}\PY{p}{,} \PY{n}{mean\PYZus{}calculated}\PY{p}{)}
            \PY{n}{sd} \PY{o}{=} \PY{n}{math}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{n}{variance\PYZus{}calculated}\PY{p}{)}
            \PY{n}{sem} \PY{o}{=} \PY{n}{sd} \PY{o}{/} \PY{n}{math}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{n+nb}{len}\PY{p}{(}\PY{n}{data}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{l+m+mi}{1}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s1}{\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{} }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{ \PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}\PYZhy{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n}{entity\PYZus{}type}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Mean: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n+nb}{round}\PY{p}{(}\PY{n}{mean\PYZus{}calculated}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Median: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n+nb}{round}\PY{p}{(}\PY{n}{median}\PY{p}{(}\PY{n}{data}\PY{p}{)}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Variance: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n+nb}{round}\PY{p}{(}\PY{n}{variance\PYZus{}calculated}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Standard Deviation: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n+nb}{round}\PY{p}{(}\PY{n}{sd}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Standard Error: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s1}{\PYZsq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n+nb}{round}\PY{p}{(}\PY{n}{sem}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)}
            
            \PY{n}{plt}\PY{o}{.}\PY{n}{hist}\PY{p}{(}\PY{n}{data}\PY{p}{,} \PY{l+m+mi}{9}\PY{p}{)}
            \PY{n}{plt}\PY{o}{.}\PY{n}{xlabel}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Recognition Time (in seconds)}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
            \PY{n}{plt}\PY{o}{.}\PY{n}{ylabel}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Number of Words}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
            \PY{n}{plt}\PY{o}{.}\PY{n}{title}\PY{p}{(}\PY{n}{entity\PYZus{}type}\PY{p}{)}
            \PY{n}{plt}\PY{o}{.}\PY{n}{show}\PY{p}{(}\PY{p}{)}
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}6}]:} \PY{n}{congruent\PYZus{}timings} \PY{o}{=} \PY{p}{[}\PY{n+nb}{float}\PY{p}{(}\PY{n}{row}\PY{p}{[}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Congruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{]}\PY{p}{)} \PY{k}{for} \PY{n}{row} \PY{o+ow}{in} \PY{n}{timings}\PY{p}{]}
        \PY{n}{describe\PYZus{}data}\PY{p}{(}\PY{n}{congruent\PYZus{}timings}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{congruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
        
        \PY{n}{incongruent\PYZus{}timings} \PY{o}{=} \PY{p}{[}\PY{n+nb}{float}\PY{p}{(}\PY{n}{row}\PY{p}{[}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Incongruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{]}\PY{p}{)} \PY{k}{for} \PY{n}{row} \PY{o+ow}{in} \PY{n}{timings}\PY{p}{]}
        \PY{n}{describe\PYZus{}data}\PY{p}{(}\PY{n}{incongruent\PYZus{}timings}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Incongruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]

-------------------------- congruent --------------------------
Mean: 14.05
Median: 14.36
Variance: 12.67
Standard Deviation: 3.56
Standard Error: 0.74

    \end{Verbatim}

    \begin{center}
    \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{output_6_1.png}
    \end{center}
    { \hspace*{\fill} \\}
    
    \begin{Verbatim}[commandchars=\\\{\}]

-------------------------- Incongruent --------------------------
Mean: 22.02
Median: 21.02
Variance: 23.01
Standard Deviation: 4.8
Standard Error: 1.0

    \end{Verbatim}

    \begin{center}
    \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{output_6_3.png}
    \end{center}
    { \hspace*{\fill} \\}
    
    \section{A picture is worth a thousand
words}\label{a-picture-is-worth-a-thousand-words}

Now that I have my measures for central tendencies, I will plot the
distribution of number of words w.r.t recognition time for both
congruent and incongruent lists.

    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}7}]:} \PY{n}{stacked\PYZus{}timings} \PY{o}{=} \PY{p}{[}\PY{n}{congruent\PYZus{}timings}\PY{p}{,} \PY{n}{incongruent\PYZus{}timings}\PY{p}{]}
        \PY{n}{label\PYZus{}texts} \PY{o}{=} \PY{p}{[}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Congruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Incongruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{]}
        \PY{n}{plt}\PY{o}{.}\PY{n}{hist}\PY{p}{(}\PY{n}{stacked\PYZus{}timings}\PY{p}{,} \PY{l+m+mi}{12}\PY{p}{,} \PY{n}{histtype}\PY{o}{=}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{bar}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{n}{stacked}\PY{o}{=}\PY{k+kc}{True}\PY{p}{,} \PY{n}{label}\PY{o}{=}\PY{n}{label\PYZus{}texts}\PY{p}{)}
        \PY{n}{plt}\PY{o}{.}\PY{n}{xlabel}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Recognition Time (in seconds)}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
        \PY{n}{plt}\PY{o}{.}\PY{n}{ylabel}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Number of Words}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
        \PY{n}{plt}\PY{o}{.}\PY{n}{title}\PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Grouped timings for Congruent and Incongruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
        \PY{n}{plt}\PY{o}{.}\PY{n}{legend}\PY{p}{(}\PY{p}{)}
        \PY{n}{plt}\PY{o}{.}\PY{n}{show}\PY{p}{(}\PY{p}{)}
\end{Verbatim}


    \begin{center}
    \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{output_8_0.png}
    \end{center}
    { \hspace*{\fill} \\}
    
    \subsection{Observations from our
graphs}\label{observations-from-our-graphs}

Quite clearly the time taken for incomgruent lists is more than that of
congruent lists. Also, both charts appear to be right skewed.

    \section{Statistical test and
Results}\label{statistical-test-and-results}

For my tests, I will use the following:

α = .01

degrees of freedom = 23

t-critical value = 2.807

    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}8}]:} \PY{n}{difference\PYZus{}data} \PY{o}{=} \PY{p}{[}\PY{p}{(}\PY{n+nb}{float}\PY{p}{(}\PY{n}{row}\PY{p}{[}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Congruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{]}\PY{p}{)} \PY{o}{\PYZhy{}} \PY{n+nb}{float}\PY{p}{(}\PY{n}{row}\PY{p}{[}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Incongruent}\PY{l+s+s1}{\PYZsq{}}\PY{p}{]}\PY{p}{)}\PY{p}{)} \PY{k}{for} \PY{n}{row} \PY{o+ow}{in} \PY{n}{timings}\PY{p}{]}
        \PY{n}{diff\PYZus{}mean} \PY{o}{=} \PY{n}{mean}\PY{p}{(}\PY{n}{difference\PYZus{}data}\PY{p}{)}
        \PY{n}{diff\PYZus{}variance} \PY{o}{=} \PY{n}{variance}\PY{p}{(}\PY{n}{difference\PYZus{}data}\PY{p}{,} \PY{n}{diff\PYZus{}mean}\PY{p}{)}
        \PY{n}{diff\PYZus{}standard\PYZus{}deviation} \PY{o}{=} \PY{n}{math}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{n}{diff\PYZus{}variance}\PY{p}{)}
        \PY{n}{t\PYZus{}stat} \PY{o}{=} \PY{n+nb}{round}\PY{p}{(}\PY{n}{diff\PYZus{}mean}\PY{o}{/}\PY{p}{(}\PY{n}{diff\PYZus{}standard\PYZus{}deviation} \PY{o}{/} \PY{n}{math}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{n+nb}{len}\PY{p}{(}\PY{n}{difference\PYZus{}data}\PY{p}{)}\PY{p}{)}\PY{p}{)}\PY{p}{,} \PY{l+m+mi}{2}\PY{p}{)}
        \PY{n+nb}{print}\PY{p}{(}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{t\PYZhy{}statistic calculated is: }\PY{l+s+si}{\PYZob{}\PYZcb{}}\PY{l+s+s2}{\PYZdq{}}\PY{o}{.}\PY{n}{format}\PY{p}{(}\PY{n}{t\PYZus{}stat}\PY{p}{)}\PY{p}{)}
\end{Verbatim}


    \begin{Verbatim}[commandchars=\\\{\}]
t-statistic calculated is: -8.02

    \end{Verbatim}

    At α = .01 or 99\% confidence level, t-critical value is 2.807 for a two
tailed test. Since t-statistic is outside the t-critical region, I will
reject the null hypothesis. I also noticed that t-statistic is negative,
so the difference between congruent and incongruent lists seems to be
negative. In other words, incongruent lists take more time to read. At α
= .01, we can say that there is a statisticaly significant difference in
time taken to recongnize the color for congruent and incongruent lists.

My initial intution was that incongruent lists would take more time to
read, but for my experiment, I just wanted to determine if there was
significant difference between both the lists, in either direction. As
we can see from the result, there is a difference and as expected.

    \section{Think some more}\label{think-some-more}

\begin{itemize}
\item
  What do you think is responsible for the effects observed?

  If I have to make a guess, I would say that human brain can either
  focus on color or the word. If the color and word are same, there is
  just one recognition task, either of color or word. When the color and
  word don't match, there is another decision that brain has to make,
  say the color or say the word. This decision making depends on each
  person and hence the time taken for incongruent lists will be more for
  most people.
\item
  Can you think of an alternative or similar task that would result in a
  similar effect?

  Another task that can be similar to this would be counting number of
  characters in word. One set of words will be words for numbers, for
  example four, three, two, ten. Another set of words would be any other
  word that doesn't represent a number. The set with numbers should be
  difficult to count as the mind would try to read the word as number
  rather than counting characters.
\end{itemize}


    % Add a bibliography block to the postdoc
    
    
    
    \end{document}