# File: mit18_05_s22_studio8-grader.r
# Authors: Jeremy Orloff and Jennifer French
#
# MIT OpenCourseWare: https://ocw.mit.edu
# 18.05 Introduction to Probability and Statistics
# Spring 2022
# For information about citing these materials or our Terms of Use, visit:
# https://ocw.mit.edu/terms.
#
# Studio 8 grading script
# Expected output in studio8-grader.html
# If this file changes --need to rebuild studio*-grader.html
# Use 'File > Compile report...' to create an R Markdown report from this.
# Because this opens a new session, it doesn't see the environment.
# So we need the following line, which should be commented out when using the grading script for grading.
source('mit18_05_s22_studio8-solutions.r') ### COMMENT OUT FOR GRADING
cat("WARNING: make sure source('mit18_05_s22_studio*-solutions.r') is commented out before grading\n")
## WARNING: make sure source('mit18_05_s22_studio*-solutions.r') is commented out before grading
# For grading, open this file and set working directory to source file location
studio8_problem_1(7, 12, 7, 3, 10000)
## ----------------------------------
## Problem 1: Simulate F statistic
## See plots
# Generated data with the following code
# printarray = function(x, name, rnd) {
# if (name != '') {
# cat(name, ' = ', sep='')
# }
# cat('c(', sep='')
# cat(round(x, rnd), sep=', ')
# cat(')\n')
# }
# n = 100
# x = rnorm(n, 5, 3)
# printarray(x, 'grade_data_problem_2', 3)
#
# n = 16
# T1 = rnorm(n, 3, 2)
# T2 = rnorm(n, 3, 2.5)
# T3 = rnorm(n, 3, 2)
# printarray(T1, 'T1', 3)
# printarray(T2, 'T2', 3)
# printarray(T3, 'T3', 3)
grade_data_problem_2 = c(12.217, 3.57, -2.703, 0.653, 5.092, 4.35,
6.39, 4.872, 6.357, 0.891, 9.489, 13.523,
2.446, 4.848, 5.061, 3.858, 10.368, 6.457,
5.032, 1.283, 9.803, 4.274, 7.705, 7.908,
1.901, 3.546, 2.906, 6.397, 7.417, 6.675,
1.7, 3.424, 4.526, 0.296, 3.12, 6.113,
5.228, 7.037, 6.992, 1.46, 1.334, 3.812,
0.72, 8.16, -0.94, 4.83, 6.861, 10.365,
2.485, 1.754, 7.712, 4.995, 2.061, 5.622,
7.864, 5.101, 8.978, 2.97, 7.046, 4.988,
6.619, 5.281, 2.7, 10.346, 7.425, 2.039,
5.084, 7.623, 10.517, 4.579, 11.71, 6.949,
0.627, 4.082, 5.864, 8.183, 3.89, 3.513,
10.302, 3.392, 6.154, 8.054, 1.368, 6.464,
7.77, 4.589, 5.961, 6.07, 9.4, 3.898,
4.242, 7.963, 4.907, 4.759, 7.903, 9.101,
10.617, 7.79, 6.306, 9.564)
studio8_problem_2(grade_data_problem_2, 5, 3, 0.1)
## ----------------------------------
## Problem 2: Code z-test by hand
## mean of data = 5.52805
## z = 1.760167
## p = 0.07837955
## Because the p-value 0.07837955 < 0.1 , the data supports rejecting the null hypothesis in favor of the alternative.
studio8_problem_2(grade_data_problem_2, 5, 3, 0.01)
## ----------------------------------
## Problem 2: Code z-test by hand
## mean of data = 5.52805
## z = 1.760167
## p = 0.07837955
## Because the p-value 0.07837955 > 0.01 , the data does not support rejecting the null hypothesis in favor of the alternative.
studio8_problem_2(grade_data_problem_2, 0, 3, 0.1)
## ----------------------------------
## Problem 2: Code z-test by hand
## mean of data = 5.52805
## z = 18.42683
## p = 0
## Because the p-value 0 < 0.1 , the data supports rejecting the null hypothesis in favor of the alternative.
studio8_problem_2(grade_data_problem_2, 0, 3, 0.01)
## ----------------------------------
## Problem 2: Code z-test by hand
## mean of data = 5.52805
## z = 18.42683
## p = 0
## Because the p-value 0 < 0.01 , the data supports rejecting the null hypothesis in favor of the alternative.
studio8_problem_3("mit18_05_s22_studio8Problem3_grade.tbl", 0.05)
## ----------------------------------
## Problem 3: Chi-square test for independence
## Married.once Married.more
## College 500 100
## No College 600 160
## ---Using chisq.test
##
## Pearson's Chi-squared test
##
## data: contingency_tbl
## X-squared = 4.1713, df = 1, p-value = 0.04112
##
## The p-value 0.04111503 is less than or equal to 0.05 , so at significance level 0.05 we should reject H0 in favor of the alternative that the levels of marriage and education are not independent.
## ----------
## Same calculation by hand.
## Test stat X^2 = 4.171267
## p-value = 0.04111503
studio8_problem_3("mit18_05_s22_studio8Problem3_grade.tbl", 0.01)
## ----------------------------------
## Problem 3: Chi-square test for independence
## Married.once Married.more
## College 500 100
## No College 600 160
## ---Using chisq.test
##
## Pearson's Chi-squared test
##
## data: contingency_tbl
## X-squared = 4.1713, df = 1, p-value = 0.04112
##
## The p-value 0.04111503 is greater than 0.01 , so at significance level 0.01 the data does not support rejecting H0, that the levels of marriage and education are independent.
## ----------
## Same calculation by hand.
## Test stat X^2 = 4.171267
## p-value = 0.04111503
T1 = c(1.592, -0.102, 3.924, 4.333, 4.68, 4.22, 2.749, -0.016,
0.16, 6.067, 2.152, 2.211, 2.86, 5.831, 3.941, 3.104)
T2 = c(2.052, -0.785, 2.24, 5.096, -1.119, 4.221, 2.687, -0.134,
8.31, 2.265, 2.169, 0.957, 3.851, 3.562, 1.218, 3.67)
T3 = c(4.812, 5.025, 3.341, 1.199, 1.588, 3.911, 5.098, 2.667,
4.152, 0.49, 3.625, 2.848, 4.15, -0.041, 2.26, -1.563)
studio8_problem_4(T1, T2, T3, 0.05)
## ----------------------------------
## Problem 4: ANOVA using aov()
## procedure pain
## 1 T1 1.592
## 2 T1 -0.102
## 3 T1 3.924
## 4 T1 4.333
## 5 T1 4.680
## 6 T1 4.220
## 7 T1 2.749
## 8 T1 -0.016
## 9 T1 0.160
## 10 T1 6.067
## 11 T1 2.152
## 12 T1 2.211
## 13 T1 2.860
## 14 T1 5.831
## 15 T1 3.941
## 16 T1 3.104
## 17 T2 2.052
## 18 T2 -0.785
## 19 T2 2.240
## 20 T2 5.096
## 21 T2 -1.119
## 22 T2 4.221
## 23 T2 2.687
## 24 T2 -0.134
## 25 T2 8.310
## 26 T2 2.265
## 27 T2 2.169
## 28 T2 0.957
## 29 T2 3.851
## 30 T2 3.562
## 31 T2 1.218
## 32 T2 3.670
## 33 T3 4.812
## 34 T3 5.025
## 35 T3 3.341
## 36 T3 1.199
## 37 T3 1.588
## 38 T3 3.911
## 39 T3 5.098
## 40 T3 2.667
## 41 T3 4.152
## 42 T3 0.490
## 43 T3 3.625
## 44 T3 2.848
## 45 T3 4.150
## 46 T3 -0.041
## 47 T3 2.260
## 48 T3 -1.563
##
## Summary:
## Df Sum Sq Mean Sq F value Pr(>F)
## procedure 2 1.74 0.870 0.2 0.819
## Residuals 45 195.37 4.341
##
## p-value: 0.8191381
## The p-value 0.8191381 is greater than 0.05 . So, at significance level 0.05 the data does not support rejecting H0, that the levels of pain in the different treatments are all the same.
studio8_problem_4(T1, T1, T1, 0.01)
## ----------------------------------
## Problem 4: ANOVA using aov()
## procedure pain
## 1 T1 1.592
## 2 T1 -0.102
## 3 T1 3.924
## 4 T1 4.333
## 5 T1 4.680
## 6 T1 4.220
## 7 T1 2.749
## 8 T1 -0.016
## 9 T1 0.160
## 10 T1 6.067
## 11 T1 2.152
## 12 T1 2.211
## 13 T1 2.860
## 14 T1 5.831
## 15 T1 3.941
## 16 T1 3.104
## 17 T2 1.592
## 18 T2 -0.102
## 19 T2 3.924
## 20 T2 4.333
## 21 T2 4.680
## 22 T2 4.220
## 23 T2 2.749
## 24 T2 -0.016
## 25 T2 0.160
## 26 T2 6.067
## 27 T2 2.152
## 28 T2 2.211
## 29 T2 2.860
## 30 T2 5.831
## 31 T2 3.941
## 32 T2 3.104
## 33 T3 1.592
## 34 T3 -0.102
## 35 T3 3.924
## 36 T3 4.333
## 37 T3 4.680
## 38 T3 4.220
## 39 T3 2.749
## 40 T3 -0.016
## 41 T3 0.160
## 42 T3 6.067
## 43 T3 2.152
## 44 T3 2.211
## 45 T3 2.860
## 46 T3 5.831
## 47 T3 3.941
## 48 T3 3.104
##
## Summary:
## Df Sum Sq Mean Sq F value Pr(>F)
## procedure 2 0.0 0.000 0 1
## Residuals 45 166.3 3.696
##
## p-value: 1
## The p-value 1 is greater than 0.01 . So, at significance level 0.01 the data does not support rejecting H0, that the levels of pain in the different treatments are all the same.