inst/tests/valid_signature_1.md

student: David K student_id: 10101010 url: http://127.0.0.1/lab_04.Rmd pubkey: MFkwEwYHKoZIzj0CAQYIKoZIzj0DAQcDQgAECEBmlRqws4b7xZSp4dhunPFrpRXbsqF9gPIsihjqOqC3lSpqnvenZmx3WD9A1+uw65vbCYfbyGIav343Y1frVg== signature: MEYCIQCtxksHFKhvneJbyCtLrALRrhdqVt5uG0DQwb5l47rOSQIhAIH4V10Me3plx7Ubci+ZXLBBHX5cBCtuB1ww69UUOoEe;MEYCIQChEz+v4AqGA4pIssu18YdbqjKvKuC7uH+FMYnzDTA4zAIhAJlR6o7MmWjOKPBaZo7xCyhkGjcpdHdCbO8xOqkey8h9

Answers

applet_idv

125

applet_n

100

applet_p1

0.15

applet_p2

0.45

applet_p3

0.1

applet_p4

0.3

applet_sim

1

q1a_1-generate_table

## Fill in the probabilities column wise.
observed_frequencies <- matrix(rmultinom(1, size = 125, prob = c(0.15, 0.45, 0.10, 0.30)), nrow = 2)
# the matrix function creates a 2 by 2 table by specifying nrow = 2
observed_frequencies

Result

     [,1] [,2]
[1,]   24   10
[2,]   50   41

q1a_2-chi_square_test

## Chose the right logical indicator
observed_frequencies
chisq.test(observed_frequencies, correct = FALSE)

Result

     [,1] [,2]
[1,]   28   13
[2,]   43   41

    Pearson's Chi-squared test

data:  observed_frequencies
X-squared = 3.2842, df = 1, p-value = 0.06995

q2a_1-condition1-31a117b8-text

123

q2a_2-condition1-discussion-3c0afda6-text

asdf

q2b_1-condition2-10ea1942-text

123

q2b_2-condition2-discussion-4312db87-text

asdf

q2c_1-condition3-3f2dc821-text

123

q2c_2-condition3-discussion-65891582-text

asdf

q2d_1-condition4-1f170046-text

123

q2d_2-condition4-discussion-50873ca4-text

asdf

zz_group-helped_most_first

d

zz_group-helped_most_last

d

zz_group-member_names

d


dakep/stat305templates documentation built on Nov. 27, 2022, 8:23 a.m.