TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW11.3

An experiment producing numerical data

Suppose we have an experiment that produces numerical data with the possible outcomes −2, −1, 0, 1, 2, 3 or more. We run 55 trials and count the frequency of each outcome, getting the following data: See Table 1.

Use the $\mathrm {X} ^{2}$-test for goodness of fit to test the null hypothesis $H_{0}$ that the data is drawn from 55 trials of a binomial distribution $B(8,0.5)$ against the alternative hypothesis $H_{A}$ that the data is drawn from some other distribution with the level of significance $\alpha =1\%$. Compute the p-value

and give your decision on the test.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Table 1

Outcomes	-2	-1	0	1	2	3 or more
Observed Frequencies	3	11	16	14	8	3

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

# Suppose we have an experiment that produces numerical data with the possible outcomes −2,
# −1, 0, 1, 2, 3 or more. We run 55 trials and count the frequency of each outcome, getting the
# following data:

data <- c("-2"=3, "-1"=11, "0"=16, "1"=14, "2"=8, "3+"=3)
n <- 55
d <- 6

barplot(data / n, col = rainbow(d))

# Use the χ^2-test for goodness of fit to test the null hypothesis H0 that the data is drawn from 55
# trials of a binomial distribution B(8, 0.5) against the alternative hypothesis HA that the data is
# drawn from some other distribution with the level of significance α = 1%. Compute the p-value
# and give your decision on the test.

alpha <- 0.01

#p_0 <- c(dbinom(2:6, 8, 0.5), pbinom(7, 8, 0.5, lower.tail = FALSE))

# p_0 <- c(dbinom(2:6, 8, 0.5), 1 - sum(dbinom(2:6, 8, 0.5)))

# == correction by tutor ==
p_0 <- c(dbinom(-2:2, 8, 0.5), pbinom(3, 8, 0.5, lower.tail = FALSE))
p_0

e <- p_0 * n
e

# here we get expected frequencies of zero because binom only works for > 0
# the hypothesised distribution cannot fit the data and we reject 

# == end correction by tutor ==

x2 <- sum((data - e)^2 / e)
x2

p <- pchisq(x2, d - 1, lower.tail = FALSE)
p

R <- c(qchisq(1 - alpha, d - 1), Inf)

p > alpha

# p > alpha therefore we don't reject the null hypothesis

?chisq.test
chisq.test(data, p=p_0)  

plot(-4:4, dbinom(0:8, size=8, prob=0.5),type='h')