def gen_ctx():
ctx = Chi2GoodnessOfFitData(seed = seed)
#Here we sample from a non-uniform distribution for the die!
ctx1_args = {
'o_dist':[1/5, 1/5, 1/5, 1/5, 1/10, 1/10],
'alternative':"The die is not fair.",
'note':"""
For this problem the truth is tha the die is not fair. \
If you accepted H<sub>0</sub>, then this is a <strong>miss</strong>
(Type II error).
"""
}
ctx1 = Chi2GoodnessOfFitData(seed = seed, **ctx1_args)
######################################
# Pass the pigs game: Due to embedding the table in the story
# this one is a little harder to set up. The problem is that this
# pollutes namespace for future contexts.
# Append _ to avoid namespace pollution
_outcomes = ['Pink', 'Dot', 'Razorback', 'Trotter',
'Snouter', 'Leaning Jowler']
_t_dist = [.35, .30, .20, .10, .04, .01]
tb = Table(_t_dist, col_headers = _outcomes,
row_headers = ['Position', 'Expected Frequency'])
if fmt == 'html':
styles = Table.get_style()
tbl = tb.html()
else:
styles = ""
tbl = tb.latex()
_s_size = random.randint(20, 30) * 10
ctx2_args = {
'outcome_type':'Position',
'outcomes':_outcomes,
't_dist':_t_dist,
's_size':_s_size,
'story':"""Pass The Pigs® is a game from Milton-Bradley™
which is essentially a dice game, except that instead of dice
players toss small plastic pigs that can land in any of 6
positions. For example, you roll a 'trotter' if the pig falls
standing on all 4 legs. It is claimed that the distribution
for the 6 positions are:
{styles}
{tbl}
To test this you toss a pig {s_size} times and get the observed
frequencies below:
""".format(styles = styles, tbl = tbl, s_size = _s_size),
'null':"The observed values of the positions of the pigs agrees \
with the expected distribution.",
'alternative':"The observed values of the positions of the pigs \
differs from what should be the case if the expected distribution\
was correct.",
'note':"""
In this case the observed data was sampled from the given
distribution. So if the null hypothesis is rejected, this
is <string>a Type-I error</strong> or <strong>a false
positive</strong>.
"""
}
ctx2 = Chi2GoodnessOfFitData(seed = seed, **ctx2_args)
###########################################
## 11.2 from text
s_size = random.randint(5, 10) * 10
ctx3_args = {
'outcomes':['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday',
'Friday', 'Saturday'],
't_dist':np.ones(7) * 1/7,
's_size':s_size,
'a_level': random.choice([0.1,0.01,0.05]),
'outcome_type':'Day of Week',
'story':"""
Teachers want to know which night each week their students are
doing most of their homework. Most teachers think that students
do homework equally throughout the week. Suppose a random sample
of %s students were asked on which night of the week they
did the most homework. The results were distributed as in
""" % s_size,
'null':"Students are equally likely to do the majority of their \
homework on any of the seven nights of the week.",
'alternative':"Students are more likely to do the majority of their \
homework on certain nights rather than others."
}
ctx3 = Chi2GoodnessOfFitData(seed = seed, **ctx3_args)
return [ctx, ctx1, ctx2, ctx3]
def explanation(self, path = 'explanations',
expanded = True, a_type = 'preview',
force = False, xkcd = False, fmt = 'latex'):
"""
Provides an explanation of the hyperbola.
Parameters: (Largely shared with show)
-----------
path : str
The output directory for image files
file_name : str
The name of output file without extension.
a_type : Strint ('MC', 'preview')
If 'preview', then set up for previewing.
fmt : String ('html', 'latex')
If 'latex' us MathJax, if 'html' use CSS tables.
"""
if fmt == 'html':
ex = Table.get_style()
else:
ex = ""
file_name = path + "/" + self.url + ".png"
if a_type == 'preview':
file_name = file_name.replace('%2','%252')
const = self.a ** 2 * self.b **2
x_part = const*(self(x, self.k).expand() - self(0, self.k))
y_part = const*(self(self.h, y).expand() - self(self.h, 0))
rhs = const - const*self(0, 0)
# The following is an attempt to avoid lots of code duplication.
if self.trans == 'x':
X = x
Y = y
X_part = x_part
Y_part = y_part
HH = self.h
KK = self.k
TX = 1
TY = 0
else:
X = y
Y = x
X_part = y_part
Y_part = x_part
HH = self.k
KK = self.h
TX = 0
TY = 1
# To reduce typing
A = self.a
B = self.b
H = self.h
K = self.k
C = self.c
k, h, a, b, c = sym.symbols('k h a b c')
gcd_X = sym.gcd(X_part.collect(X).coeff(X,1), X_part.collect(X).coeff(X,2))
gcd_Y = sym.gcd(Y_part.collect(Y).coeff(Y,1), Y_part.collect(Y).coeff(Y,2))
b_X = X_part.collect(X).coeff(X,1)/gcd_X
b_Y = -Y_part.collect(Y).coeff(Y,1)/gcd_Y
if expanded:
ex += "The first step to finding the graph of $$%s = 0$$ is\
to find the normal form of the equation." \
% sym.latex((const*(self(x, y) - 1)).expand())
ex += " To begin move the constant term to the right hand side. \
This gives: $$(%s) + (%s) = %s$$" \
% (sym.latex(x_part), sym.latex(y_part), rhs)
ex += "Next factor common factors from the $_x$_ and $_y$_ terms to get: \
$$%s(%s) - %s(%s) = %s$$" \
% (gcd_X if gcd_X != 1 else "", sym.latex(X_part/gcd_X),
gcd_Y if gcd_Y != 1 else "", sym.latex(-Y_part/gcd_Y), rhs)
ex += "Now complete the squares: \
$$%s\\left(%s %s\\right) \
- %s\\left(%s %s\\right) = \
%s %s %s$$"\
% (gcd_X if gcd_X != 1 else "", sym.latex(X_part/gcd_X),
"+ \\left(\\frac{%s}{2}\\right)^2" % b_X if b_X != 0 else "",
gcd_Y if gcd_Y != 1 else "", sym.latex(-Y_part/gcd_Y),
"+ \\left(\\frac{%s}{2}\\right)^2" % b_Y if b_Y != 0 else "",
rhs,
"+ %s\\left(\\frac{%s}{2}\\right)^2" % (gcd_X, b_X) if b_X != 0 else "",
"- %s\\left(\\frac{%s}{2}\\right)^2" % (gcd_Y, b_Y) if b_Y != 0 else "")
ex += "This simplifies to: $$%s\\left(%s\\right)^2 - %s\\left(%s\\right)^2 = %s$$" \
% (gcd_X if gcd_X != 1 else "",
sym.latex(X + b_X/2),
gcd_Y if gcd_Y != 1 else "",
sym.latex(Y + b_Y/2),
const)
ex += "Lastly divide both side by the right hand side to get: \
$$%s\\left(%s\\right)^2 - %s\\left(%s\\right)^2 = 1$$" \
% (sym.latex(gcd_X/const) if gcd_X/const != 1 else "",
sym.latex(X + b_X/2),
sym.latex(gcd_Y/const) if gcd_Y/const != 1 else "",
sym.latex(Y + b_Y/2))
ex += "This simplifies to the final normal form: \
$$\\frac{(%s)^2}{%s^2} - \\frac{(%s)^2}{%s^2} = 1$$" \
% (sym.latex(X-HH), A, sym.latex(Y-KK), B)
else:
ex += "The hyperbola is given in standard normal form: \
$$\\frac{(%s)^2}{%s^2} - \\frac{(%s)^2}{%s^2} = 1$$" \
% (sym.latex(X-HH), A, sym.latex(Y-KK), B)
ex += "From this we can read off the center to be at $_(h,k) = (%s, %s)$_. "\
% (H,K)
ex += "The tansverse (major) axis is along $_%s = %s$_ and has length $_2a = %s$_. "\
% (sym.latex(Y), K, 2*A)
ex += "The vertices are $_(%s, %s) = (%s,%s)$_ and $_(%s, %s) = (%s, %s)$_. "\
% (sym.latex(h - a * TX), sym.latex(k - a * TY), H - A * TX, K - A * TY,
sym.latex(h + a * TX), sym.latex(k + a * TY), H + A * TX, K + A * TY)
ex += "The conjugate (minor) axis is along $_%s = %s$_ and has length \
$_2b = %s$_. " % (sym.latex(Y), H, 2*B)
ex += "The co-vertices are $_(%s, %s) = (%s,%s)$_ and $_(%s, %s) = (%s, %s)$_. "\
% (sym.latex(h - B * (1 - TX)), sym.latex(k - B * (1 - TY)),
H - B * (1 - TX), K - B * (1 - TY),
sym.latex(h + B * (1 - TX)), sym.latex(k + B * (1 - TY)),
H + B * (1 - TX), K + B * (1 - TY))
ex += "The two assymptotes are $_y = \\pm\\frac{b}{a}(%s) %s %s \
= \\pm\\frac{%s}{%s}(%s) %s %s $_. " \
% (sym.latex(x - H), "+" if K > 0 else "-",
sym.Abs(K), B, A, sym.latex(x - H), "+" if K > 0 else "-",
sym.Abs(K))
ex += "Finally, the focal length is $_c = \\sqrt{a^2+b^2}=%s$_ and the foci \
are located at $_(%s, %s) = (%s,%s)$_ and $_(%s, %s) = (%s, %s)$_. "\
% (sym.latex(C),
sym.latex(h - c * TX), sym.latex(k - c * TY),
sym.latex(H - C * TX), sym.latex(K - C * TY),
sym.latex(h + c * TX), sym.latex(k + c * TY),
sym.latex(H + C * TX), sym.latex(K + C * TY))
data = [
['center', '$_(%s, %s)$_' % (H, K)],
['vertices', '$_(%s, %s), (%s, %s)$_' \
% tuple(map(sym.latex, [H - A * TX, K - A * TY,
H + A * TX, K + A *TY]))],
['length of conjugate axis', '$_%s$_' % sym.latex(2*A)],
['co-vertices', '$_(%s, %s), (%s, %s)$_' \
% tuple(map(sym.latex, [H - B * (1 - TX), K - B * (1 - TY),
H + B * (1 - TX), K - B * (1 - TY)]))],
['length of conjugate axis', '$_%s$_'% sym.latex(2*B)],
['foci', '$_(%s, %s), (%s, %s)$_' \
% tuple(map(sym.latex, [H - C * TX, K - C * TY,
H + C * TX, K + C * TY]))],
['asymptotes', '$_y = \\pm\\frac{%s}{%s}(%s) %s %s$_' \
% tuple(map(sym.latex, [B, A, sym.sympify(x - H),
"+" if K > 0 else "-",
sym.Abs(K)]))]
]
tb = Table(data)
if fmt == 'latex':
tbl = tb.latex()
else:
tbl = tb.html()
img = html_image(image_url = file_name, width = '300px',
preview = (a_type == 'preview'))
ex +="""
%s
<div class='outer-container-rk'>
<div class='centering-rk'>
<div class='container-rk'>
<figure>
%s
<figcaption>$_%s = 1$_</figcaption>
</figure>
</div>
<div class='container-rk'>
%s
</div>
</div>
</div>
""" % (Table.get_style(), img, self.latex, tbl)
self.show(path = path, file_name = self.url, force = force,
xkcd = xkcd)
return ex
def stem(self, context = None, table = None, q_type = None,
a_type = 'preview', force = False, fmt = 'html'):
"""
This will generate a problem for $\chi^2$ goodness of fit.
Parameters:
----------
context : context object
This describes the problem. A default context is used if this is
none.
table : string ['hist', 'table']
Display t_dist as a table (html/latex) or as a histogram.
q_type : string [None, 'STAT', 'HT', 'CI']
If None randomly choose. If 'STAT' just compute the chi2 statistic
and the degrees of freedom. If 'HT, compute the p-value for the
data and determine whether or not reject the null hypothesis. If
'CI' compute the confidence interval.
a_type : string
This is eithe "MC" or "preview" for now
fmt : String ['html', 'latex']
Use nice CSS/HTML (responsive) or plain LaTeX (static)
Notes:
-----
The default here is to simulate a roll of a die 30 times and record the
frequency of values. The :math`\chi^2` test should test whether the die
is fair at an :math:`alpha`
level of 5%.
"""
kwargs = {
'context': context,
'table': table,
'q_type': q_type,
'fmt': fmt,
'a_type': a_type
}
if q_type is None:
q_type = random.choice(['STAT', 'HT', 'PVAL'])
if table == None:
table = random.choice(['table', 'hist'])
if context == None:
context = Chi2GoodnessOfFitData()
if not context.is_valid:
warnings.warn("Context had invalid cell counts.")
return
# Generate unique name
q_name = hash(context)
self.cache.add(q_name)
self.hist = str(q_name) + "_hist.png"
self.solution_plot = str(q_name) + "_plot.png"
style = None
if a_type == 'preview':
question_stem = "<h2>Question</h2><br>"
else:
question_stem = ""
if fmt == 'html':
question_stem += Table.get_style()
question_stem += "<div class='par'>" + context.story + "</div>\n"
if table == 'table':
if fmt == 'html':
tbl = context.observed.html()
question_stem += tbl
else:
tbl = context.observed.latex()
question_stem += tbl
elif table == 'hist':
fname = context.hist(path = self.path, force = force)
img = html_image(fname, width = '300px',
preview = (a_type == 'preview'))
if style is None:
style = Table.get_style()
question_stem += style
question_stem +="""
<div class='outer-container-rk'>
<div class='centering-rk'>
<div class='container-rk'>
<figure>
%s
<figcaption>%s</figcaption>
</figure>
</div>
</div>
</div>
""" % (img, "Observed Frequencies")
N = len(context.outcomes)
df = N - 1
chi2eq = "$$\\chi^2_{%s}=\\sum_{i=1}^{%s}\\frac{(O_i-E_i)^2}{E_i}\
= %.3g$$" % (df, N, context.chi2_stat)
if q_type == 'STAT':
question_stem += "Compute the $_\\chi^2$_-statistic and degrees \
of freedom for the given observed values."
elif q_type == 'PVAL':
question_stem += """The $_\\chi^2$_ statistic is
{chi2eq}
Use this information to find the degrees of freedom (df) and the
$_p\\text{{-value}} = P(\\chi^2_{{{df}}} > {chi2:.3g})$_.
""".format(chi2eq=chi2eq, N=N, df = 'df', chi2=context.chi2_stat)
elif q_type == 'HT':
question_stem += """The degrees of freedom are $_df = {N} - 1 =
{df}$_ and the $_\\chi^2$_ statistic is {chi2eq}
Use this information to conduct a hypothesis test with
$_\\alpha = {a_level}$_. Choose the answer that best captures
the null hypothesis and conclusion.
""".format(N = N, df = df, chi2eq = chi2eq,
a_level = context.a_level)
if fmt == 'html':
explanation = Table.get_style()
else:
explanation = ""
if a_type == 'preview':
explanation += "<br><h2>Explanation</h2><br>"
tb1 = Table(context.t_dist, col_headers = context.outcomes,
row_headers = [context.outcome_type, 'Probabilities'])
if fmt == 'html':
tbl1 = tb1.html()
tbl2 = context.oe.html()
else:
tbl1 = tb1.latex()
tbl2 = context.oe.latex()
explanation += "<div class='par'>To find the expected counts multiply the total\
number of observations by the expected probability for an outcome. \
The probabilities for the expected outcomes are summarized in the \
following table:"
explanation += tbl1
explanation += "and there are %s observations." % context.s_size
explanation += " So the expected and observed counts are:<br>"
explanation += tbl2
explanation += "</div>"
explanation += "<div class='par'>The degrees of freedom are $_df = {N} - 1 = \
{df}$_ and the $_\\chi^2$_-statistic is:{chi2eq}</div>"\
.format(N=N,df=df,chi2eq=chi2eq)
if q_type in ['HT','PVAL']:
rv = stats.chi2(df)
p_val = 1 - rv.cdf(context.chi2_stat)
fname = context.show(path = self.path, force = force)
img = html_image(fname, width = '300px',
preview = (a_type == 'preview'))
caption = """
Lightshading (right of the red line) indicates the \p-value.
<br>
The darker shading indicate the $_\\alpha = $_ {a_level:.0%}
level.<br>
The background histogram is a bootstrap sampling distribution.
""".format(a_level = context.a_level)
explanation +="""
The p-value for this data is:
$$\\text{p-value} = P(\\chi^2_{%s} > %.3g) = %.4g%s$$
<div class='outer-container-rk'>
<div class='centering-rk'>
<div class='container-rk'>
<figure>
%s
<figcaption>%s</figcaption>
</figure>
</div>
</div>
</div>
""" % (df, context.chi2_stat, p_val * 100, '\\%', img, caption)
if q_type == 'HT':
if p_val < context.a_level:
explanation += """
<div class='par'>The p-value is less than the
$_\\alpha$_-level so the null hypothesis is rejected.
That is, we accept the alternative hypothesis:
<strong>H<sub>a</sub>: {alt} </strong>
More precisely, assuming the null hypothesis, there
is only a {p_val:.2%} probability due to
random chance in sampling that
the difference in the expected and observed data is
least this large.</div>
""".format(alt=context.alternative, p_val=p_val)
else:
explanation += """
<div class='par'>The p-value is greater than the $_\\alpha$_-level so
the null hypothesis
is not rejected. Precisely, assuming the null hypothesis
<strong>H<sub>0</sub>: {null}</strong>,
there is a {p_val:.2%} probability due to
random chance in sampling that
the difference in the expected and observed data is at
least this large.</div>
""".format(null=context.null, p_val=p_val)
explanation += """
<div class='par'>Note: {note}</div>
""".format(note=context.note)
errors = self.gen_errors(q_type, context)
if a_type == 'preview':
errs = [[er] for er in errors]
choices = "<br><h2>Choices</h2><br>"
tb = Table(errs, row_headers = ['Answer'] + ['Distractor']*4)
# Choices need the richer html structure in preview.
choices += Table.get_style() + tb.html()
if fmt == 'html':
return question_stem + choices + explanation
else:
return question_stem.replace("<div ","<p ")\
.replace("</div>","</p>") + choices + \
explanation.replace("<div ","<p ")\
.replace("</div>","</p>")
elif a_type == 'MC':
if fmt == 'latex':
question_stem = question_stem.replace("<div ","<p ")\
.replace("</div>","</p>")
explanation = explanation.replace("<div ","<p ")\
.replace("</div>","</p>")
distractors = [err.replace("<div ","<p ")\
.replace("</div>","</p>") for err in errors]
question_stem = ' '.join(question_stem.split())
distractors = [' '.join(err.split()) for err in errors]
explanation = ' '.join(explanation.split()) + "\n"
return tools.fully_formatted_question(question_stem, explanation,
answer_choices=distractors)
elif a_type == 'Match':
pass
else:
pass