Fitting probability models to frequency data (Part 2)
M. Drew LaMar
October 16, 2019
Class Announcements
Chi-squared goodness-of-fit test
Working through an example
Assignment Problem #21
A more recent study of Feline High-Rise Syndrom (FHRS) included data on the month in which each of 119 cats fell (Vnuk et al. 2004). The data are in the accompanying table. Can we infer that the rate of cat falling varies between months of the year?
| Month | Number fallen | Month | Number fallen |
|---|---|---|---|
| January | 4 | July | 19 |
| February | 6 | August | 13 |
| March | 8 | September | 12 |
| April | 10 | October | 12 |
| May | 9 | November | 7 |
| June | 14 | December | 5 |
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Example - Assignment Problem #21
A more recent study of Feline High-Rise Syndrom (FHRS) included data on the month in which each of 119 cats fell (Vnuk et al. 2004). The data are in the accompanying table. Can we infer that the rate of cat falling varies between months of the year?
Question: What are the null and alternative hypotheses?
Answer:
\( H_{0} \): The frequency of cats falling is the same in each month.
\( H_{A} \): The frequency of cats falling isnot the same in each month.
And then an R miracle occurred
And then an R miracle occurred
Example - Assignment Problem #21
mytable %>%
select(-month) %>%
as.matrix() %>%
t() %>%
barplot(beside = TRUE,
col = c("forestgreen",
"goldenrod1"),
legend.text = c("Observed",
"Expected"),
names.arg = mytable$month)
Example - Assignment Problem #21
tl;dr
Example - Assignment Problem #21
Example - Assignment Problem #21
Definition: The
\( \chi^2 \) statistic measures the discrepancy between observed frequencies from the data and expected frequencies from the null hypothesis and is given by
\[ \chi^2 = \sum_{i}\frac{(Observed_{i} - Expected_{i})^2}{Expected_{i}} \]
Example - Assignment Problem #21
Definition: The
\( \chi^2 \) statistic measures the discrepancy between observed frequencies from the data and expected frequencies from the null hypothesis and is given by
\[ \chi^2 = \sum_{i}\frac{(Observed_{i} - Expected_{i})^2}{Expected_{i}} \]
Discuss: What would support the null hypothesis more: a small value or large value for \( \chi^{2} \)?
Answer: Small value for \( \chi^{2} \)
Example - Assignment Problem #21
\( \chi^2 \) test statistic - computed in R
(mytable %>%
mutate(tmp = ((obs-exp)^2)/exp) %>%
summarize(chi2 = sum(tmp)))
# A tibble: 1 x 1
chi2
<dbl>
1 20.7
1x1 tibble?! Here's where “everything is a data frame” doesn't work.
Solution?
Example - Assignment Problem #21
magrittr package
“The Treachery of Images” by René Magritte
Example - Assignment Problem #21
\( \chi^2 \) test statistic - computed in R
Use the magrittr package and the exposition pipe operator %$%.
mytable %>%
mutate(tmp = ((obs-exp)^2)/exp) %>%
summarize(chi2 = sum(tmp)) %$%
chi2
[1] 20.66387
The observed \( \chi^2 \) test statistic is 20.6638655.
Is that good?
Example - Assignment Problem #21
Example - Assignment Problem #21
Definition: The number of
degrees of freedom of a \( \chi^2 \) statistic specifies which \( \chi^2 \) distribution to use as the null distribution and is given by
df = Number of categories - 1 - Number of estimated parameters from data
This is analogous to the sample size in the null distribution for a mean and proportion. Remember, for the mean and proportion, the null distribution depends on the sample size.
Example - Assignment Problem #21
\( P \)-value = 0.0370255 (shaded red area)
Example - Assignment Problem #21
A more recent study of Feline High-Rise Syndrom (FHRS) included data on the month in which each of 119 cats fell (Vnuk et al. 2004). The data are in the accompanying table. Can we infer that the rate of cat falling varies between months of the year?
\( P \)-value = 0.0370255
Discuss: Conclusion?
Conclusion: Reject \( H_{0} \), i.e. there is evidence that the frequency of cats falling is
not the same in each month.
Example - Assignment Problem #21
Critical value = \( \chi_{0.05,11}^2 = 19.6751376 \)
Red area = \( \alpha = 0.05 \)
Example - Assignment Problem #21
Statistical tables
\[ Pr[\chi^{2} \geq \chi_{0.05,11}^2] = Pr[\chi^{2} \geq 19.675] = 0.05 \] \[ \mathrm{Observed} \ \chi^2 = 20.6638655 \]
Important - Assumptions to use test
Which probability models for this chapter?
Example - Practice Problem #6 (Binomial)
Example - Practice Problem #6 (Binomial)
Example - Practice Problem #6 (Binomial)
Example - Practice Problem #6 (Binomial)
Example - Practice Problem #6 (Binomial)
Example - Practice Problem #6 (Binomial)
Statistical tables
\[ df = 7 - 1 = 6 \] \[ Pr[\chi^{2} \geq \chi_{0.05,6}^2] = Pr[\chi^{2} \geq 12.59] = 0.05 \] \[ \mathrm{Observed} \ \chi^2 = 302.3218794 \]