W5Q1: Find a percentile

Description ----------------------------------------------------------

The |Statistics Toolbox| makes available a number of high-level
functions, |prctile| being one of them. Note that, to compute a 50th
percentile, the function |median| would do just fine (and does not
require the Toolbox!).

Starting today, I hope, the |Statistics Toolbox| should be enabled for
your set of weekly problems.

*A vector contains all the integers 1 through 100. A same-length
 vector contains numbers that are powers, equally spaced between 0 and
 1. Calculate the value of the 37th, the 86th, and the 95th
 percentiles of the elements of the first vector raised to the powers
 of the second (in an element-by-element sense), and assign the
 results to the vector variable* |vp|.

Solution Template ----------------------------------------------------

% Define the percentiles ahead of time
prc=
% Calculate the requested percentile for the requested data set
vp=

Reference Solution ---------------------------------------------------

prc=[37 86 95];
vp=prctile([1:100].^linspace(0,1,length(v)),prc);

Visible Tests --------------------------------------------------------


Hidden Tests ---------------------------------------------------------

%%
run('solution')
assert(isequal(vw,[1:100].^linspace(0,1,100)))
assessVariableEqual('vp',prctile(vw,prc),'AbsoluteTolerance',1e-10)


W5Q2: Judge a correlation

Description ----------------------------------------------------------

*Use the function* |corr| *to calculate the correlation coefficient
 between the two variables of a given data set*, |x1| *and* |x2|.

*When properly used, the* _first_ *output of* |corr| *is a simple
 number, the correlation coefficient that we are after. Collect this
 value in a variable* |r|. *The* _second_ *output of* |corr| *is
 another number, with the result of a* _very specific_ *significance
 test in. Collect this value in a variable* |p|. *It is to be
 interpreted as the probability that a correlation as high as
 observed, or even higher, might arise by chance if the variables
 whose correlation you are testing are _normally_ distributed (which
 they may not be, and which should be independently tested!) and
 _uncorrelated_ (that is the null hypothesis). This very particular*
 _p-value_ *needs to be very low, say, below 5%, for us to accept that
 the variables in question are _correlated_ (which is, for us to
 _reject_ the null hypothesis). Collect the result of the test using
 the p-value at this 95% significance level (does* |p| *fall below the
 5% threshold?) in a logical variable that you name* |t|.*

_Note: unlike for_ |corrcoef|, _the_ |Statistics Toolbox| _is required
 for_ |corr|, _and it should be available under Cody. You may want to
 compare the behavior of inputs and outputs of these two functions._

Solution Template ----------------------------------------------------

% Here are the data, do not touch the two lines below
x1=[ 1  5 -6  8  9  2 0 -2 18 29 54];
x2=[-1  4 -6  4 12 -2 3 -2 33 19 47];
% Here is the significance level
alfa=0.95;
% Here is what we want to know about the data
[r,p]=
% Here you conduct the test
t=

Reference Solution ---------------------------------------------------

x1=[ 1  5 -6  8  9  2 0 -2 18 29 54];
x2=[-1  4 -6  4 12 -2 3 -2 33 19 47];
[r,p]=corr(x1(:),x2(:));
alfa=0.95;
t=p<(1-alfa);

Visible Tests --------------------------------------------------------


Hidden Tests ---------------------------------------------------------

%%
run('solution')
assessVariableEqual('r',0.9277,'AbsoluteTolerance',1e-4)
assessVariableEqual('p',3.8644e-05,'AbsoluteTolerance',1e-9)
assessVariableEqual('t',logical(1))


W5Q3: Perform date conversions

Description ----------------------------------------------------------

_MATLAB invented its own serial date format, as you can read about
 using |help datenum|. Wrap your heads around it! You will need to go
 back and forth often._

*A certain date and time are contained in the
 numeric variable* |bd1|. *Convert this number using* |datestr| *into
 a date string that evaluates to the format* |dd-mm-yyyy|, *and call
 your new variable* |bds1|.

*A certain other date are giving to you as a string, in a variable
 named* |bds2|. *Convert this date to a numeric MATLAB date number,
 and store the result in a variable that you name* |bd2|.

*Assuming a year has 365 days and 12 months, what is the difference
 between those dates, rounded to years? And that same difference, but
 instead, rounded to months? And that same difference, rounded to
 days? Call your results* |yearswiser|, |monthswiser|, *and*
 |dayswiser|.

_Draw your own conclusions, but keep them to yourselves._


Solution Template ----------------------------------------------------

% The first date, as a MATLAB date number
bd1=721206;
% Convert it to a string in the requested format
bds1=
% The second date, as a user-generated date string
bds2='Dec-30-1975';
% Convert it to a MATLAB date number
bd2=
% How much time between those dates to various rounded calendar units?
yearswiser=
monthswiser=
dayswiser=

Reference Solution ---------------------------------------------------

bd1=721206;
bds1=datestr(bd1,'dd-mm-yyyy');
bds2='Dec-30-1975';
bd2=datenum(bds2,'mmm-dd-yyyy');
yearswiser=round([bd2-bd1]/365);
monthswiser=round([bd2-bd1]/365*12);
dayswiser=round([bd2-bd1]);

Visible Tests --------------------------------------------------------


Hidden Tests ---------------------------------------------------------

%% 
run('solution')
assert(isequal(bds1,datestr(721206,'dd-mm-yyyy')))
assert(isequal(bd2,datenum('Dec-30-1975','mmm-dd-yyyy')))
assert(isequal(yearswiser,round([bd2-bd1]/365)))
assert(isequal(monthswiser,round([bd2-bd1]/365*12)))
assert(isequal(dayswiser,round([bd2-bd1])))


W5Q4: Generate a sine wave

Description ----------------------------------------------------------

*A sequence of hours is contained in a vector* |h|. *Generate a sine
 wave that comprises the superposition of three specific periods* |P|
 *with three specific amplitudes* |A|, *oscillating about a specific
 mean value* |m|. *Collect the resulting superposition of sine waves
 in the vector* |v|.

Solution Template ----------------------------------------------------

% Here are the hours being considered, do not change these
h=0:1:100;
% Here are the three amplitudes, do not change these
A=[1 2 3];
% Here are the three periods, in the units of h, do not change
P=[6 12 24];
% Here is the mean, do not change it
m=13.5;
% Now generate the superposition of sine waves
v=

Reference Solution ---------------------------------------------------

h=0:1:100;
A=[1 2 3];
P=[6 12 24];
m=13.5;
v=m+A(:)'*sin(2*pi*h./P(:));

Visible Tests --------------------------------------------------------


Hidden Tests ---------------------------------------------------------

%% 
run('solution')
assessVariableEqual('v',m+A(:)'*sin(2*pi*h./P(:)),'AbsoluteTolerance',1e-6)


W5Q5: Fourier transform

Description ----------------------------------------------------------

*A certain sequence of data is given to you as a set of independent
 and dependent variables. Using* |fft|, *perform a calculation to
 obtain the _periodogram_ at a set of* |nfft| *frequencies, and use
 the result to determine the* _period_ *at which the sequence shows
 the largest-amplitude periodicity. Collect that answer in the
 variable* |mper|. 

_Verify that your answer makes sense!_

Solution Template ----------------------------------------------------

% The sampling interval of a sequence, do not change
xsint=1;
% A certain sequence, independent variable, do not change
h=1:xsint:6897;
% A certain sequence, dependent variable, do not change
v=[3 2 1]*sin(2*pi./[19 ; 33 ; 11]*h);
% The number of points in the Fourier transform, do not change
nfft=length(h);
% The frequency axis, do not change
selekt=[1:nfft/2+1];
fax=[selekt-1]/xsint/nfft;
% The overcomplete periodogram at nfft frequencies (not windowed)
vf=
% The periodogram at the frequencies that matter, do not change
vf=vf(selekt);
% The index of the frequency with maximum periodogram value
[~,ind]=
% Use the index to determine the period corresponding to the maximum
mper=;

Reference Solution ---------------------------------------------------

xsint=1;
h=1:xsint:6897;
v=[3 2 1]*sin(2*pi*h./[19 ; 33 ; 11]);
nfft=length(h);
selekt=[1:nfft/2+1]; 
fax=[selekt-1]/xsint/nfft;
vf=abs(fft(v,nfft)).^2;
vf=vf(selekt);
[~,faxi]=max(vf);
mper=1/fax(faxi);

Visible Tests --------------------------------------------------------


Hidden Tests ---------------------------------------------------------

%% 
run('solution')
assessVariableEqual('mper',19,'AbsoluteTolerance',1e-1)