Midterm
Matthew Martino
knitr::opts_chunk$set(echo = TRUE)
knitr::opts_chunk$set(error = TRUE)Due: 3/1/2024 @ 12:30 PM
Total points: 83
This midterm is open note but not “open internet”. You can reference
your notes, class slides, class tutorials, homeworks, and textbook to
complete these questions. You can also reference R documentation using
the ? function. Everything else is off-limits. You CANNOT
work on this with other students, ask chatGPT for help, or use Google,
Stack Overflow, or other online resources. We take academic integrity
very seriously and will report violations through the appropriate
channels.
Write your name below to acknowledge these instructions. You must do so in order to receive a grade for this exam.
Matthew Martino
Collaboration Question: Did you collaborate with other students, access resources external to class, or otherwise receive help on this assignment?
No
Part 1: Conceptual Questions (2 pts each)
For multiple choice questions, indicate your selection by bolding it. For example:
- This is an example question.
- The wrong answer
- The right answer is bold
- The wrong answer
- The wrong answer
- The phrase “all models are wrong but some are useful” most closely
addresses which of the following?
- That statistics is overly complicated.
- That statistics is both an art and a science.
- That statistics can illustrate a simplification of reality.
- That statistics connects scientific theories to the world.
- What type of information exemplifies descriptive statistics?
- Mean
- Variance
- Skew
- All of the above
- None of the above
- What does the median represent?
- The difference between the lowest and higher value in a distribution.
- The value that divides a distribution exactly in half.
- The value that occurs most often in a distribution
- An average calculated by summing all the values in a distribution, and then dividing by the number of values in the distribution.
- What is the interquartile range (IQR)?
- The difference between the third and first quartile.
- The difference between the fourth and first quartile.
- The average difference between quantiles.
- The difference between the second and first quartile.
- What type of visualization makes the most sense when displaying the
IQR?
- Bar chart
- Histogram
- Boxplot
- Scatterplot
- The law of total probability is the sum of probabilities of what?
- A series of random phenomena
- The sample space
- Repeated events
- All continuous events
- What is the probability of getting a score that is less than 1
standard deviation above the mean given a gaussian distribution?
- .16
- .50
- .68
- .84
- The spread of the sampling distribution for the sample mean is
mainly determined by which magnitude?
- The sample size
- The type I error α
- The sample mean
- The population size
- The positively skewed distribution of raw scores will result in what
type of distribution of sample means when n is sufficiently large?
- Positively skewed
- Negatively skewed
- Normally distributed
- Kurtotic
- Which of the following is the best reason for why we make an
adjustment to the formula when estimating the population standard
deviation?
- Because our sample standard deviation is a bit larger than the population standard deviation.
- The original formula would produce an unbiased estimation.
- The new formula would produce a biased estimation.
- The original formula would produce a biased estimation.
- Which of the following is the best measure of the amount of sampling
error associated with a sample mean?
- Population standard deviation
- Sample standard deviation
- Standard deviation estimate
- Standard error of the mean
- As the degrees of freedom associated with an estimate of a
population value increase, the accuracy of the estimate tends to
- Decrease
- Increase
- Stay the same
- Sometimes decrease and sometimes increase
- Which of the following is one way you might define a p-value?
- The probability that the null hypothesis is true
- The probability that the alternative hypothesis is true
- The probability that a score is more extreme than our test statistic, given H0 is true
- The probability that a score is more extreme than our test statistic, given HA is true
- Say that we select samples of size 100 from the IQ scores
population, which has a mean 110 and standard deviation 15. What would
the mean of the sampling distribution of the sample mean be?
- 95 or 125 (i.e. 110±15)
- 110
- 11
- 100
- CMU has a total enrollment of around 14000 students. Say we would
like to estimate the average SAT score of the people that are finally
accepted in CMU. To do that, each of the 100 brilliant students from
several statistics classes across campus (some of them could be you)
decides to take random samples of 50 students from the CMU population
and construct a 99% confidence interval for μ, the average SAT score of
the 14000 CMU students (i.e. the population mean). Which of the
following conclusions is correct?
- About 99% of the confidence intervals will contain the true population mean μ.
- About 1% of the sample means will not be included in the confidence intervals.
- About 1% of the individual SAT scores in the samples will not be found in these confidence intervals.
- 99% of the confidence intervals will be identical.
- A researcher believes that college graduates are generally more
satisfied with their lives than the general population. In order to
investigate this issue, the researcher administers the Satisfaction With
Life (SWL) questionnaire to a random sample of 75 college graduates.
Scores on the SWL range from 0 (not at all satisfied) to 30 (very
satisfied). The mean SWL score for the general population is known to be
20. In the sample, the mean SWL score was 23 and the standard deviation
was 7. This researcher wants to run a statistical test to find out
whether this statement is supported by the data. The researcher uses an
alpha level of .01 in order to provide additional protection against a
Type 1 error. What would their null hypothesis, H0, and their
alternative hypothesis, Ha, be?
- H0:μ < 20, HA:μ > 20
- H0:μ > 23, HA:μ ≤ 23
- H0:μ = 23, HA:μ > 23
- H0:μ = 20, HA:μ > 20
- Which of the following statements best describes what a Type I error
α is in the previous question?
- 1.0% of the times that we accept the null, we will be correct.
- 1.0% of the times that we reject the null, we will be mistaken.
- 1.0% of the times that we accept the null, we will be mistaken.
- 1.0% of the times that we reject the null, we will be correct.
- The researcher determines the critical value to be 2.39 against
their test statistic. Should the researcher reject or fail to reject the
null hypothesis? You’ll need to calculate the test statistic on your own
here.
- Reject H0
- Fail to reject H0
- Irrespective of your answer to the above question, if the researcher
were to reject the null hypothesis, what would the study implications
be?
- Graduating college is related to increased life satisfaction.
- Graduating college is not related to increased life satisfaction.
- Graduating college is related to increased life satisfaction in 1% of college graduates.
- Not graduating college is associated with increased life satisfaction.
- You are going to run a study for which you expect the effect to
observe to be small. If you decide to increase the sample size of the
study:
- The size of the effect will increase
- The size of the effect will decrease
- The chance of detecting the effect will decrease
- The chance of detecting the effect will increase
Part 2: Applied Questions
Question 1 (3 pts):
Write a function that calculates the mode of a vector. If there are multiple modes, just return one of them (it doesn’t matter which one.)
Hints:
- The function
max()can be used to pull the maximum value from a set of numbers. - We recently learned a function that quickly assesses frequencies…
myMode <- function(v){
# helper code: This code removes NAs from your vector
# and returns NA if the vector is empty
v <- v[!is.na(v)]
if(length(v) == 0){return(NA)}
## start your code below:
orderedTable <-
table(v) %>%
sort( decreasing = TRUE)
orderedTable[1] %>%
names() %>%
return()
}You can test your function on the code below (ungraded):
modeTest <- c("a", "a", "a", "a", "a", "a",
"b", "b",
"c", "c", "c", "c", "c", "c", "c", "c",
"d", "d", "d")
#uncomment to test
myMode(modeTest)## Error in table(v) %>% sort(decreasing = TRUE): could not find function "%>%"Question 2: Getting to know your data (13 pts)
This next section will use data collected at Carleton University by Bradley & Howard (2022). Over 12 weeks (during 2020), the authors surveyed college students about their mood and collected screen time and app data from their phones. You do not have to read the published article by these authors. We’ll only use a portion of the variables.
Download the data file from Canvas to your 309 folder, and load the data below:
library(tidyverse)## Warning: package 'tidyverse' was built under R version 4.3.2## Warning: package 'ggplot2' was built under R version 4.3.2## Warning: package 'tibble' was built under R version 4.3.2## Warning: package 'tidyr' was built under R version 4.3.2## Warning: package 'readr' was built under R version 4.3.2## Warning: package 'purrr' was built under R version 4.3.2## Warning: package 'dplyr' was built under R version 4.3.2## Warning: package 'stringr' was built under R version 4.3.2## Warning: package 'forcats' was built under R version 4.3.2## Warning: package 'lubridate' was built under R version 4.3.2## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.5.0 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.1
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors# load your data here:
screen.data <- read.csv("screenData.csv")A. Use the appropriate function to view the internal structure of the data. (1 pt)
# your answer here
structure(screen.data)## id week screenDailyHrs topAppDailyHrs avgstress topApp weekHappy
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## 2 1 2 6.683333 1.8357143 0.0000000 socMedia NA
## 3 1 3 7.783333 2.0309524 0.0000000 socMedia 2
## 4 1 4 6.383333 1.7690476 0.7500000 socMedia 1
## 5 1 5 5.983333 1.7190476 0.0000000 entertainment 1
## 6 1 6 8.433333 3.6309524 0.0000000 entertainment 3
## 7 1 7 5.316667 1.7619048 0.0000000 socMedia 5
## 8 1 8 7.433333 2.7809524 0.0000000 entertainment 4
## 9 1 9 5.516667 1.8166667 1.3333333 socMedia 1
## 10 1 10 4.850000 1.2404762 0.0000000 socMedia 3
## 11 1 11 NA 2.2261905 0.0000000 socMedia 2
## 12 1 12 5.933333 2.5000000 0.0000000 socMedia 1
## 13 2 2 6.466667 2.5452381 2.3333333 socMedia 3
## 14 2 6 6.150000 1.4976190 2.0000000 socMedia 3
## 15 3 2 7.300000 1.6095238 0.7500000 entertainment 2
## 16 3 3 NA NA 0.8444444 <NA> 4
## 17 3 4 7.150000 1.5095238 2.0000000 socMedia 2
## 18 3 5 5.116667 0.9214286 0.0000000 socMedia 2
## 19 3 12 7.050000 1.4500000 0.7500000 socMedia 3
## 20 4 2 4.683333 1.2666667 3.7500000 socMedia NA
## 21 4 3 3.966667 1.4166667 1.5000000 socMedia 3
## 22 4 4 4.550000 1.8619048 2.0000000 socMedia 3
## 23 4 5 5.300000 1.7238095 0.0000000 socMedia 4
## 24 4 6 4.616667 1.9976190 0.0000000 socMedia 4
## 25 4 7 5.233333 1.4833333 0.7500000 socMedia 2
## 26 4 8 3.883333 1.4928571 0.0000000 socMedia 4
## 27 4 9 4.450000 1.4666667 0.7500000 socMedia 4
## 28 5 2 NA NA 1.2045455 <NA> 3
## 29 5 5 3.416667 NA 2.2500000 <NA> 2
## 30 5 6 4.933333 NA 4.0000000 <NA> 3
## 31 5 9 NA 2.4619048 3.0000000 entertainment 3
## 32 5 11 3.600000 NA 2.7500000 <NA> 4
## 33 6 2 7.316667 5.1333333 2.7500000 entertainment NA
## 34 6 3 NA 4.0333333 3.2500000 entertainment 3
## 35 6 4 6.650000 4.6976190 3.2500000 entertainment 2
## 36 6 5 7.400000 4.5523810 3.7500000 entertainment 1
## 37 6 6 8.533333 5.8452381 2.7500000 entertainment 2
## 38 6 7 7.983333 4.7809524 2.0000000 entertainment 4
## 39 6 8 7.800000 5.1142857 2.5000000 entertainment 2
## 40 6 9 7.416667 4.2738095 2.7500000 entertainment 2
## 41 6 10 8.750000 5.0714286 1.7500000 entertainment 2
## 42 6 11 9.783333 5.9952381 1.7500000 entertainment 2
## 43 6 12 12.550000 8.7809524 2.7500000 entertainment 1
## 44 7 2 7.383333 1.3880952 0.7500000 socMedia 4
## 45 7 3 9.283333 1.8976190 2.7500000 socMedia 5
## 46 7 4 9.166667 2.1619048 4.0000000 socMedia 2
## 47 7 5 8.033333 1.3666667 3.0000000 socMedia 4
## 48 7 6 6.733333 1.7690476 3.7500000 socMedia 4
## 49 7 7 8.233333 1.6523810 3.0000000 socMedia 3
## 50 7 10 NA NA 1.8750000 <NA> 2
## 51 7 11 5.483333 1.2666667 2.7500000 socMedia 5
## 52 8 3 4.800000 2.1785714 1.0000000 socMedia 2
## 53 8 4 7.016667 3.6833333 3.3333333 socMedia 5
## 54 8 5 7.066667 3.0714286 1.3333333 socMedia 3
## 55 8 6 6.683333 3.2642857 0.0000000 socMedia 4
## 56 8 8 6.533333 2.9833333 0.0000000 socMedia 3
## 57 8 9 4.483333 1.4214286 0.0000000 socMedia 1
## 58 8 10 7.666667 4.2833333 2.6666667 socMedia 2
## 59 9 4 10.050000 1.4690476 0.0000000 socMedia 4
## 60 9 5 12.850000 1.6119048 0.7500000 entertainment 4
## 61 9 6 6.700000 1.2666667 0.0000000 information 5
## 62 10 7 NA NA 1.2131148 <NA> 3
## 63 10 8 NA NA 1.3469388 <NA> 3
## 64 11 1 4.683333 1.8428571 3.7500000 socMedia 1
## 65 11 2 5.200000 2.2595238 2.7500000 socMedia 1
## 66 11 5 4.283333 2.2142857 2.0000000 socMedia 1
## 67 11 6 5.333333 1.7238095 0.7500000 socMedia 1
## 68 11 10 4.350000 1.8166667 4.0000000 socMedia 1
## 69 12 1 2.333333 0.2904762 3.5000000 other 4
## 70 13 1 2.666667 NA 1.2500000 <NA> 3
## 71 13 2 2.800000 NA 0.0000000 <NA> 3
## 72 13 3 3.150000 1.7904762 1.2500000 socMedia 3
## 73 13 4 2.650000 1.2261905 1.2500000 socMedia 2
## 74 13 5 2.816667 1.4428571 0.5000000 socMedia 3
## 75 13 6 2.866667 1.3904762 1.2500000 socMedia 2
## 76 13 7 2.850000 1.2833333 2.7500000 socMedia 3
## 77 13 8 2.800000 1.4071429 1.7500000 socMedia 2
## 78 13 9 2.733333 0.9666667 1.7500000 socMedia 3
## 79 13 10 2.483333 0.9761905 2.0000000 socMedia 3
## 80 13 11 2.216667 1.1047619 1.7500000 socMedia 3
## 81 13 12 2.566667 1.5380952 1.7500000 socMedia 2
## 82 14 1 5.583333 1.6738095 0.0000000 socMedia 1
## 83 14 2 4.783333 2.3071429 1.5000000 socMedia 1
## 84 14 3 NA NA 0.8444444 <NA> 2
## 85 14 4 NA NA 1.1095890 <NA> 1
## 86 14 5 NA NA 1.5769231 <NA> 2
## 87 14 6 NA NA 1.3030303 <NA> 1
## 88 14 7 NA NA 1.2131148 <NA> 1
## 89 14 8 NA NA 1.3469388 <NA> 1
## 90 14 9 10.750000 4.4690476 0.0000000 entertainment 1
## 91 14 10 11.833333 3.8261905 1.0000000 entertainment 1
## 92 14 11 10.500000 3.8000000 2.0000000 entertainment 1
## 93 14 12 9.200000 2.0666667 0.0000000 entertainment 1
## 94 15 1 NA NA 1.3103448 <NA> 4
## 95 16 2 4.366667 1.2261905 0.7500000 socMedia 4
## 96 16 3 9.466667 5.9190476 0.7500000 entertainment 3
## 97 16 4 9.116667 4.9619048 1.7500000 entertainment 2
## 98 16 5 10.316667 5.8880952 0.0000000 entertainment 3
## 99 16 6 10.083333 4.8523810 0.7500000 entertainment 1
## 100 16 7 NA NA 1.2131148 <NA> 3
## 101 16 8 NA NA 1.3469388 <NA> 4
## 102 16 9 NA NA 1.4888889 <NA> 5
## 103 16 10 NA NA 1.8750000 <NA> 4
## 104 16 11 6.150000 2.4000000 0.0000000 socMedia 3
## 105 16 12 NA NA 0.4285714 <NA> 4
## 106 17 2 5.050000 1.2857143 1.7500000 socMedia 5
## 107 17 4 3.900000 1.0380952 1.7500000 socMedia 4
## 108 17 5 4.516667 1.3166667 1.7500000 socMedia 2
## 109 17 6 4.550000 1.4904762 1.0000000 socMedia 5
## 110 17 7 5.233333 1.3500000 3.5000000 socMedia 4
## 111 17 8 4.983333 1.0095238 1.0000000 socMedia 2
## 112 17 10 5.033333 1.5690476 0.0000000 entertainment 4
## 113 17 11 5.683333 1.7976190 0.7500000 entertainment 3
## 114 17 12 4.450000 1.0761905 1.0000000 socMedia 4
## 115 18 2 5.666667 NA 2.0000000 <NA> 2
## 116 18 3 8.250000 NA 2.0000000 <NA> 2
## 117 18 4 NA NA 1.1095890 <NA> 2
## 118 18 5 8.683333 3.2071429 1.3333333 entertainment 2
## 119 18 6 8.766667 4.4238095 2.3333333 entertainment 4
## 120 18 7 7.850000 3.0142857 1.0000000 entertainment 4
## 121 18 8 7.183333 2.2952381 1.0000000 entertainment 3
## 122 18 9 7.733333 2.4714286 1.0000000 entertainment 2
## 123 18 10 8.683333 1.8666667 1.3333333 entertainment 2
## 124 18 11 7.950000 1.1452381 1.3333333 socMedia 2
## 125 18 12 8.633333 1.9428571 1.3333333 entertainment 2
## 126 19 2 5.316667 1.7952381 1.5000000 information 4
## 127 19 3 NA NA 0.8444444 <NA> 4
## 128 19 4 4.883333 1.3833333 0.0000000 information 4
## 129 19 5 5.283333 1.2166667 1.5000000 information 4
## 130 19 6 4.883333 0.8261905 1.7500000 information 3
## 131 19 7 6.616667 1.7404762 0.0000000 information 3
## 132 19 8 5.233333 1.4595238 1.0000000 information 3
## 133 19 9 4.666667 1.0928571 1.0000000 information 3
## 134 19 10 5.016667 0.6690476 1.7500000 information 4
## 135 19 11 5.066667 1.2119048 1.0000000 information 4
## 136 19 12 5.766667 0.8333333 1.0000000 socMedia 3
## 137 20 2 4.966667 0.9023810 0.0000000 socMedia 4
## 138 20 3 1.850000 0.3547619 0.0000000 information 4
## 139 20 4 2.850000 0.8714286 0.0000000 socMedia 5
## 140 20 5 4.266667 0.6738095 0.0000000 socMedia 3
## 141 20 6 4.900000 0.8833333 0.0000000 information 4
## 142 20 7 4.083333 0.6095238 0.0000000 socMedia 3
## 143 20 8 3.200000 0.5023810 0.0000000 socMedia 4
## 144 20 9 4.616667 0.9404762 0.0000000 socMedia 3
## 145 20 10 4.000000 0.9166667 0.0000000 socMedia 3
## 146 20 12 3.200000 0.7285714 0.0000000 socMedia 2
## 147 21 2 NA NA 1.2045455 <NA> NA
## 148 22 2 5.250000 1.4142857 1.7500000 socMedia 3
## 149 22 4 4.600000 1.0285714 2.0000000 socMedia 1
## 150 22 5 6.200000 1.7476190 0.0000000 socMedia 4
## 151 22 6 5.750000 2.1714286 0.0000000 socMedia 3
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## 153 22 9 7.183333 1.5976190 2.0000000 socMedia 2
## 154 22 10 8.150000 2.0785714 1.3333333 socMedia 4
## 155 23 2 NA NA 1.2045455 <NA> 2
## 156 23 4 NA NA 1.1095890 <NA> 3
## 157 24 3 4.283333 0.8119048 1.0000000 socMedia 4
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## 159 24 5 3.733333 0.7476190 0.0000000 socMedia 5
## 160 24 6 3.833333 0.7452381 0.0000000 socMedia 5
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## 174 26 4 NA NA 1.1095890 <NA> 5
## 175 26 5 17.016667 6.4142857 1.7500000 entertainment 1
## 176 27 4 4.700000 1.1666667 1.7500000 socMedia 2
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## 182 27 10 5.866667 1.3523810 2.3333333 socMedia 2
## 183 27 11 8.033333 1.8071429 2.3333333 socMedia 2
## 184 27 12 9.983333 2.2000000 2.0000000 entertainment 3
## 185 28 6 NA NA 1.3030303 <NA> 4
## 186 28 7 NA NA 1.2131148 <NA> NA
## 187 28 8 NA NA 1.3469388 <NA> 3
## 188 28 9 NA NA 1.4888889 <NA> 4
## 189 28 11 3.833333 0.4452381 0.0000000 socMedia 4
## 190 29 1 7.283333 1.6976190 1.0000000 socMedia 4
## 191 29 5 6.550000 2.6428571 1.3333333 socMedia 3
## 192 29 10 6.983333 3.6619048 0.0000000 socMedia 4
## 193 29 11 7.583333 2.9928571 0.7500000 socMedia 3
## 194 29 12 6.283333 2.2428571 0.0000000 socMedia 2
## 195 30 1 NA NA 1.3103448 <NA> 1
## 196 30 3 NA NA 0.8444444 <NA> 3
## 197 30 4 NA NA 1.1095890 <NA> 4
## 198 30 5 NA NA 1.5769231 <NA> 2
## 199 30 6 NA NA 1.3030303 <NA> 2
## 200 30 7 NA NA 1.2131148 <NA> 2
## 201 30 8 NA NA 1.3469388 <NA> 2
## 202 30 9 NA NA 1.4888889 <NA> 2
## 203 30 10 NA NA 1.8750000 <NA> 3
## 204 30 11 NA NA 1.4615385 <NA> 4
## 205 30 12 NA NA 0.4285714 <NA> 1
## 206 31 1 8.900000 2.5571429 1.5000000 socMedia 3
## 207 31 2 7.983333 2.5476190 0.0000000 socMedia 3
## 208 31 3 NA NA 0.8444444 <NA> 3
## 209 31 4 6.866667 2.1238095 2.5000000 socMedia 3
## 210 31 5 7.583333 2.5357143 0.7500000 socMedia 2
## 211 31 6 8.883333 2.5380952 1.5000000 socMedia 4
## 212 31 8 8.316667 1.8214286 1.0000000 socMedia 4
## 213 31 9 9.350000 3.0928571 0.0000000 socMedia 3
## 214 31 10 9.483333 2.6595238 0.0000000 socMedia 3
## 215 31 11 10.033333 3.1261905 0.7500000 socMedia 4
## 216 31 12 9.550000 2.6619048 0.0000000 socMedia 5
## 217 32 1 7.266667 4.5285714 0.0000000 entertainment 2
## 218 32 3 5.866667 2.8285714 0.0000000 entertainment 3
## 219 32 4 5.833333 1.6595238 0.0000000 entertainment 3
## 220 32 5 7.483333 5.1404762 1.0000000 entertainment 2
## 221 32 6 7.666667 4.1761905 0.0000000 entertainment 3
## 222 32 7 7.666667 4.1761905 1.3333333 entertainment 2
## 223 32 8 8.633333 4.1476190 0.0000000 entertainment 4
## 224 32 9 6.066667 3.0047619 0.0000000 entertainment 4
## 225 32 10 7.650000 4.3190476 0.0000000 entertainment 3
## 226 32 11 8.116667 4.6452381 0.0000000 entertainment 4
## 227 32 12 8.600000 5.6285714 1.0000000 entertainment 2
## 228 33 1 11.316667 3.8166667 1.5000000 entertainment 5
## 229 33 2 8.900000 1.7309524 1.5000000 entertainment 4
## 230 33 3 11.783333 5.6642857 2.2500000 entertainment 3
## 231 33 4 11.666667 5.8642857 3.0000000 entertainment 4
## 232 33 5 11.550000 5.4880952 2.3333333 entertainment 4
## 233 33 6 10.033333 2.5833333 2.0000000 entertainment 4
## 234 33 8 6.816667 1.1380952 0.0000000 socMedia 5
## 235 34 1 2.766667 0.6285714 0.0000000 socMedia 4
## 236 34 4 2.616667 0.8619048 1.3333333 entertainment 4
## 237 34 5 2.433333 0.6690476 1.3333333 entertainment 4
## 238 34 6 1.866667 0.3714286 3.6666667 socMedia 3
## 239 34 7 1.316667 0.2642857 2.3333333 information 3
## 240 34 8 1.716667 0.2738095 1.0000000 entertainment 3
## 241 34 9 1.316667 0.2619048 2.3333333 entertainment 2
## 242 34 10 2.250000 0.3523810 1.3333333 socMedia 2
## 243 34 11 2.833333 0.6238095 1.3333333 information 3
## 244 34 12 2.966667 0.7428571 1.0000000 information 2
## 245 35 1 4.516667 1.4928571 1.0000000 entertainment 2
## 246 35 3 3.716667 0.7333333 1.0000000 socMedia 1
## 247 35 4 5.666667 1.0142857 1.3333333 entertainment 1
## 248 35 5 3.950000 0.7000000 1.3333333 entertainment 3
## 249 35 6 4.283333 0.8119048 1.3333333 information 3
## 250 35 7 5.133333 1.4023810 1.0000000 entertainment 3
## 251 35 8 5.633333 1.3261905 0.0000000 entertainment 3
## 252 36 2 7.833333 NA 2.2500000 <NA> 3
## 253 36 4 12.616667 7.4976190 3.0000000 entertainment 3
## 254 36 5 11.383333 5.1857143 2.2500000 entertainment 2
## 255 36 8 12.983333 8.7880952 2.5000000 entertainment 2
## 256 36 10 12.766667 5.7690476 3.5000000 entertainment 2
## 257 36 11 12.166667 3.2380952 3.0000000 entertainment 2
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## 351 1
## 352 5
## 353 5
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## 372 1
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## 414 2
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## 416 2
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## 419 1
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## 425 1
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## 432 4
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## 434 4
## 435 4
## 436 2
## 437 2
## 438 2
## 439 2
## 440 1
## 441 2
## 442 2
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## 444 2
## 445 4
## 446 2
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## 448 1
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## 454 2
## 455 2
## 456 1
## 457 5
## 458 4
## 459 5
## 460 5
## 461 5
## 462 3
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## 464 5
## 465 3
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## 469 4
## 470 5
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## 472 NA
## 473 2
## 474 2
## 475 1
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## 477 2
## 478 5
## 479 1
## 480 3
## 481 1
## 482 2
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## 484 1
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## 494 4
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## 504 3
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## 506 NA
## 507 2
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## 510 5
## 511 5
## 512 5
## 513 3
## 514 3
## 515 1
## 516 1
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## 518 1
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## 537 4
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## 543 1
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## 581 2
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## 584 1
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## 587 1
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## 593 3
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## 616 3
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## 658 4
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## 660 1
## 661 1
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## 706 1
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## 708 1
## 709 1
## 710 1
## 711 1
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## 715 NA
## 716 1
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## 719 3
## 720 4
## 721 4
## 722 1
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## 724 5
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## 973 4The variables in this dataset include:
- id: Participant ID
- week: Week, 1-12
- screenDailyHrs: Daily screen hours for the given week
- topAppDailyHrs: Daily hours spent on the participants most used apps, averaged for the given week
- topApp: Category of participants’ most used app.
- weekHappy: Participants’ rating of happy mood over the week, on a scale of 1-5.
- weekStress: Participants’ rating of stress over the week, on a scale of 1-5.
B. Given this variable information and data structure, what variable(s) need to be changed to a different data type? Justify your answer. (1 pt)
topApp should be a factor because it’s categorical data
Now, conduct the data transformation(s) you specified (1 pt):
topApp <- as.factor(screen.data$topApp)C. Use code to show the categories of topApp, and the
frequency of each category. (1 pt)
table(screen.data$topApp)##
## entertainment information other socMedia
## 191 43 15 548D. What is the most common category for topApp? (1
pt)
socMedia
E. Are these data in long or wide format? (1 pt)
long
F. In the original manuscript for these data, the authors conducted time series analyses which we haven’t learned. Instead, we’ll look at these data in aggregate, i.e., averaged across the 12 weeks for each individual.
Make the following data transformations using the tidyverse pipe. (2 pts)
- Within each individual, create the following
summary statistics by aggregating over the individuals’ scores:
- mHappy: mean of weekHappy
- mStress: mean of weekStress
- mScreenHrs: mean of screenDailyHrs
- mTopAppHrs: mean of topAppDailyHrs
- modeTopApp: mode of topApp (use the function you just wrote!)
- Drop rows with missing values
- Save your modified dataframe as
aggregateData
Don’t create temporary objects to accomplish this. Be sure to use the pipe!
This should result in a data frame with 157 rows: one for each participant.
aggregateData <- screen.data %>%
drop_na() %>%
group_by(id) %>%
summarize(mHappy = mean(weekHappy),
mStress = mean(weekStress),
mScreenHrs = mean(screenDailyHrs),
mTopAppHrs = mean(topAppDailyHrs),
modeTopApp = myMode(topApp))Show the first few lines of your new data object:
head(aggregateData)## # A tibble: 6 × 6
## id mHappy mStress mScreenHrs mTopAppHrs modeTopApp
## <int> <dbl> <dbl> <dbl> <dbl> <chr>
## 1 1 2.33 2.44 6.40 2.14 socMedia
## 2 2 3 2.5 6.31 2.02 socMedia
## 3 3 2.25 3.5 6.65 1.37 socMedia
## 4 4 3.43 2.57 4.57 1.63 socMedia
## 5 6 2 4.67 8.54 5.46 entertainment
## 6 7 3.86 2.71 7.76 1.64 socMediaG. We’re now working with summary statistics for each individual across the 12 weeks. How many participants averaged less than 2 hours of screen time per day over this period? How many reported more than 6? (1 pt)
aggregateData %>%
filter(mScreenHrs < 2) %>%
nrow()## [1] 1aggregateData %>%
filter(mScreenHrs > 6) %>%
nrow()## [1] 95H. Is the average daily hours spent on participants’ top app across
the semester (mTopAppHrs) skewed? Use both histograms and a
summary statistic to answer this question.
Plot (1 pt):
ggplot(aggregateData, aes(x = mTopAppHrs)) + geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Statistics (1 pt):
summary(aggregateData$mTopAppHrs)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1976 1.3467 1.9211 2.2357 2.7926 6.9270Interpret your outputs. If relevant, be sure to indicate the direction of skew. (1pt):
Both visually and by comparing the median and mean (mean noticeably greater than median), it seems that mTopAppHrs is positively skewed.
I. Let’s visualize how screen time and stress are related to each other. (2 pts)
Using the tidyverse pipe, do the following:
- Start with your aggregateData object
- Drop the rows where the
modeTopAppis ‘other’ - Plot the data with
mStresson the x-axis andmScreenHrson the y-axis - Plot using geom_point()
- Color the points based on the participants’
modeTopApp - Add linear trend lines for each value of
modeTopApp(should be 3 total lines)
aggregateData %>%
filter(modeTopApp != 'other') %>%
ggplot(aes(x = mStress, y = mScreenHrs)) + geom_point(aes(color = modeTopApp)) + geom_smooth(method = 'lm', aes(color = modeTopApp))## `geom_smooth()` using formula = 'y ~ x'
Question 3: Sampling (9 pts)
For this question, we’ll treat our dataset
(aggregateData) as if it were a population. We’ll focus on
the mTopAppHrs variable. We’ll see how the kurtosis of the
mean sampling distribution changes with the number of samples drawn.
A. First, write two functions:
- The first should take a random sample of size n from
mTopAppHrsand calculate the sample mean - The second function should repeat the first function “samples” times (i.e., you should be able to adjust the number of samples drawn) to create a sampling distribution. It should output a single value: the kurtosis of the sampling distribution. (No need to write a function for kurtosis, you can use the one from the tutorials.)
It’s okay to adapt the code for these functions from prior work we have done in class (just double and triple check that you update all object names where necessary!)
Write your two functions here (4 pts):
library(psych)## Warning: package 'psych' was built under R version 4.3.2##
## Attaching package: 'psych'## The following objects are masked from 'package:ggplot2':
##
## %+%, alphagetSampleMean <- function(n)
{aggregateData$mTopAppHrs %>%
sample(size = n, replace = TRUE) %>%
mean()
}
varySamplesAndN_M <- function(n, samples)
{data.frame(sample = 1: samples,
sampleMeans = replicate(samples,
getSampleMean(n))) %>%
summarise(distributionMean = mean(sampleMeans)) %>%
pull()}Test your function by updating this code. It should output a single kurtosis value:
set.seed(309)
varySamplesAndN_M(n = 5, samples = 10)## [1] 2.200393B. Now, use your function to see how the kurtosis of the sampling distribution changes when the number of samples changes. Iteratively run your function with sample number ranging from 50 to 3000 by units of 10. Keep sample size (n) constant at 100. Save each output in a results data frame. (2 pts)
set.seed(309)
kurtData <- data.frame(results = numeric())
for (x in 50:3001:10){
tempKurtosis <- (varySamplesAndN_M(n = 100, samples = x))
kurtData$results <- c(results, (varySamplesAndN_M(n = 100, samples = x)))
}## Warning in 50:3001:10: numerical expression has 2952 elements: only the first
## used## Error in eval(expr, envir, enclos): object 'results' not foundC. Plot your results data frame with number of samples on the x-axis and kurtosis on the y-axis. (1 pt)
# your answer hereD. Interpret your plot. What value do the points converge on? How does this relate to the central limit theorem? (2 pts)
Write answer here
Question 4: Hypothesis Testing (10 pts)
For this question, we’ll treat aggregateData as a sample
– which is what it really is.
A. First, divide the dataset into two smaller datasets: One for people who report above-average happiness (happyHi), and one for people who report below-average happiness (happyLo). How many people fall into each group? (2 pts)
happyHi <- filter(aggregateData, mHappy < mean(aggregateData$mHappy))
happyLo <- filter(aggregateData, mHappy > mean(aggregateData$mHappy))B. Calculate the mean and 90% confidence intervals for
mScreenHrs in both groups (happyHi and happyLo). Be sure to
print out all the values (mean and upper/lower limits). (2 pts)
xBar <- mean(happyHi$mScreenHrs)
sigma <- sd(happyHi$mScreenHrs)
N <- nrow(happyHi)
t.test(x = happyHi$mScreenHrs, mu = xBar, alternative = "two.sided", conf.level = .9)##
## One Sample t-test
##
## data: happyHi$mScreenHrs
## t = 0, df = 72, p-value = 1
## alternative hypothesis: true mean is not equal to 7.17677
## 90 percent confidence interval:
## 6.611726 7.741814
## sample estimates:
## mean of x
## 7.17677 xBar <- mean(happyLo$mScreenHrs)
sigma <- sd(happyLo$mScreenHrs)
N <- nrow(happyLo)
t.test(x = happyHi$mScreenHrs, mu = xBar, alternative = "two.sided", conf.level = .9)##
## One Sample t-test
##
## data: happyHi$mScreenHrs
## t = 0.97713, df = 72, p-value = 0.3318
## alternative hypothesis: true mean is not equal to 6.845422
## 90 percent confidence interval:
## 6.611726 7.741814
## sample estimates:
## mean of x
## 7.17677C. Suppose we know that the average screen time for this population is 6.6 hours per day. (I fully made this number up.) What statistical test would be appropriate to see if each of our groups (happyHi and happyLo) were significantly different from this value?
(To clarify, this is not comparing happyHi and happyLo to each other in a single test; it is conducting 2 tests, comparing each group to the overall population.)
Justify your answer. (1 pt)
t-test because population standard deviation is unknown
D. Write the null and alternative hypothesis for each of these two tests in formal notation (4 hypotheses total). (No need to format the subscripts.) (2 pts)
nulls: happyHi screen time = 6.6, happyLo screen time = 6.6. alternatives: happyHi screen time != 6.6, happyLo screen time != 6.6
E. Conduct the two tests using an alpha level of .05. Use the built-in test function (i.e., no need to manually calculate your test statistic, critical value, etc.) (2 pts):
t.test(happyLo$mScreenHrs, alternative = "two.sided", mu = 6.6, conf.level = .95)##
## One Sample t-test
##
## data: happyLo$mScreenHrs
## t = 0.88406, df = 78, p-value = 0.3794
## alternative hypothesis: true mean is not equal to 6.6
## 95 percent confidence interval:
## 6.292748 7.398096
## sample estimates:
## mean of x
## 6.845422t.test(happyHi$mScreenHrs, alternative = "two.sided", mu = 6.6, conf.level = .95)##
## One Sample t-test
##
## data: happyHi$mScreenHrs
## t = 1.7009, df = 72, p-value = 0.09328
## alternative hypothesis: true mean is not equal to 6.6
## 95 percent confidence interval:
## 6.500782 7.852758
## sample estimates:
## mean of x
## 7.17677F. Interpret your output. What are the study implications of these tests? In other words, what did we learn here? (1 pt)
Having above and below average happiness levels significantly change the amount of screen time per week.
Question 5: \(\chi^2\) Tests (8 pts)
A. Write a function that calculates the \(\chi^2\) statistic for a frequency table. Your comparison/expected frequency (\(E_{i}\)) should be equal frequencies across all categories. For example, if I have 6 categories and 40 observations, I expect 40/6 = 6.67 observations in each category. (3 pts)
Hints:
- Try starting by calculating E. Since our null is equal frequencies, \(E_{i}\) is the same across all values of i.
- The frequency tables output by the
table()function still work element-wise, like the data frames we’ve been working with. For example, if you domyTable + 1, it will add 1 to each frequency value stored.
\[ \chi^2 = \sum_{i = 1}^k\frac{(O_{i} - E_{i})^2}{E_{i}} \]
myChi <- function(freq_table) {
# your answer here
}Use these test data to test your function. Feel free to check your work by comparing your output to the built-in function for chi-squared tests (ungraded).
testVector <- c("a", "a", "a", "a", "a", "a", "a", "a",
"b", "b",
"c", "c", "c", "c", "c", "c", "c", "c",
"d", "d", "d")
testTab <- table(testVector)
#Uncomment and update this code to test your function
#myChi(testTab)
#you can compare your calculated X^2 to the built-in function
#chisq.test(testTab)B. Uncomment and run this code to select a sample from the original
screenData dataset called appData:
#Don't alter this code
# set.seed(1)
# appSample <- sample(screenData$topApp, size = 50, replace = T)
# appSample <- appSample[!is.na(appSample)]Using your function, calculate the X^2 statistics for the
appSample object created above. Note that this object is a
vector. (1 pt)
# your answer hereC. Use the appropriate (built-in) functions to get (1) the critical X^2 value for alpha = .05, and (2) the p-value associated with the X^2 value for this sample (calculated in part B). (2 pts)
# your answer hereD. What conclusion do you draw from these results? (1 pt)
Write your answer here.
E. Check your answer against the built-in R function for X^2 tests. Did you get the same answer? (1 pt)
# your answer here