Intervals again

Question 1

At the end of last week’s exercises, we estimated the mean sleep-quality rating, and computed a confidence interval, using the formula below.

\[ \begin{align} \text{95\% CI: }& \bar x \pm 1.96 \times SE \\ \end{align} \]

Can you use R to show where the 1.96 comes from?

Hints

qnorm! (see the end of 03A #uncertainty-due-to-sampling)

Question 2

As we learned in 3B #t-distributions, the sampling distribution of a statistic has heavier tails the smaller the size of the sample it is derived from. In practice, we are better using $t$-distributions to construct confidence intervals and perform statistical tests.

The code below creates a dataframe that contains the number of books read by 7 people in 2023.
(Note tibble is just a tidyverse version of data.frame):

bookdata <- 
  tibble(
    person = c("Martin","Umberto","Monica","Emma","Josiah","Dan","Aja"),
    books_read = c(12,19,9,11,8,28,13)
  )

Calculate the mean number of books read in 2023, and construct an appropriate 95% confidence interval.

Question 3

Will a 90% confidence interval be wider or narrow?
Calculate it and see.

Procrastination Scores

Research Question Do Edinburgh University students report endorsing procrastination less than the norm?

The Procrastination Assessment Scale for Students (PASS) was designed to assess how individuals approach decision situations, specifically the tendency of individuals to postpone decisions (see Solomon & Rothblum, 1984). The PASS assesses the prevalence of procrastination in six areas: writing a paper; studying for an exam; keeping up with reading; administrative tasks; attending meetings; and performing general tasks. For a measure of total endorsement of procrastination, responses to 18 questions (each measured on a 1-5 scale) are summed together, providing a single score for each participant (range 0 to 90). The mean score from Solomon & Rothblum, 1984 was 33.

A student administers the PASS to 20 students from Edinburgh University.
The data are available at https://uoepsy.github.io/data/pass_scores.csv.

Question 4

Read in the data
Calculate some relevant descriptive statistics
Check the assumptions that we will be concerned with for a one-sample test of whether the mean PASS scores is less than 33.

Hints

What counts as “relevant statistics”? Think about the question. It concerns just one variable (the PASS scores), which is numeric. What’s a nice way of describing the center and spread of such a variable?

Question 5

Our test here is going to be have the following hypotheses:

Null: mean PASS score in Edinburgh Uni students is $\geq 33$
Alternative: mean PASS score in Edinburgh Uni students is $< 33$

Manually calculate the relevant test statistic.
Note, we’re doing this manually right now as it’s a useful learning process. In later questions we will switch to the easy way!

Hints

you can see the step-by-step calculation of a one sample t-test in 3B #one-sample-t-test.
The relevant formula is:
\[ \begin{align} & t = \frac{\bar x - \mu_0}{\frac{s}{\sqrt{n}}} \\ \qquad \\ & \text{Where:} \\ & \bar x : \text{mean of PASS in our sample} \\ & \mu_0 : \text{hypothesised mean score of 33} \\ & s : \text{standard deviation of PASS in our sample} \\ & n : \text{number of observations} \end{align} \]

Question 6

Using the test statistic calculated in question 1, compute the p-value.

Hints

this will be needing the pt() function.
the degrees of freedom is $n-1$ (we used one up by estimating the mean).
The test we are performing is against the null hypothesis that the mean is $\geq 33$. Our t-statistic is in the broad sense calculated as “mean minus 33”, so negative numbers mean we have a mean lower than 33. These are the instances that we will reject the null hypothesis - if we get a test statistic very low. So we want the lower.tail of the distribution for our p-value.

Question 7

Now using the t.test() function, conduct the same test. Check that the numbers match with your step-by-step calculations in the previous two questions.

Hints

Check out the help page for t.test() - there is an argument in the function that allows us to easily change between whether our alternative hypothesis is “less than”, “greater than” or “not equal to”.

Question 8

Create a visualisation to illustrate the results.

Question 9

Write up the results.

Hints

There are some quick example write-ups for each test in 3B #basic-tests

Heights

Research Question Is the average height of University of Edinburgh Psychology students different from 165cm?

Data: Past Surveys
In the last few years, we have asked students of the statistics courses in the Psychology department to fill out a little survey.
Anonymised data are available at https://uoepsy.github.io/data/surveydata_historical.csv.

Note: this does not contain the responses from this year.

surveydata <- 
  read_csv("https://uoepsy.github.io/data/surveydata_historical.csv")

Question 10

No more manual calculations of test statistics and p-values for this week.

Conduct a one sample $t$-test to evaluate whether the average height of UoE psychology students in the last few years was different from 165cm.

Make sure to consider the assumptions of the test!

Hints

This is real data, and real data is rarely normal! If you conduct a Shapiro-Wilk test, you may well find $p<.05$ and conclude that your data is not normal.
So what do we do if a test indicates our assumptions are violated?
Well, we should bear a couple of things in mind.
1. A decision rule such as $p<.05$ on Shapiro-Wilk test creates very dichotomous thinking for something which is in reality not black and white. Real life distributions are not either normal or non-normal. Plot the data, and make a judgement!
2. As it happens, the t-test is actually reasonably robust against slight deviations from normality, especially if the sample size is “large enough” (rule-of-thumb n = 30) and the data are not heavily skewed. Plot your data and make a judgement!

Figure 2: The deeper you get into statistics, the more you discover that it is not simply a case of following step-by-step rules.

Data

We’ll read in our data and check the dimensions and variable names

surveydata <- read_csv("https://uoepsy.github.io/data/surveydata_historical.csv")

dim(surveydata)

[1] 548  24

names(surveydata)

 [1] "birthmonth"        "height"            "eyecolour"        
 [4] "catdog"            "threewords"        "year"             
 [7] "course"            "in_uk"             "gender"           
[10] "optimism"          "spirituality"      "ampm"             
[13] "sleeprating"       "extraversion"      "agreeableness"    
[16] "conscientiousness" "emot_stability"    "imagination"      
[19] "loc"               "phone_unlocks"     "caffeine"         
[22] "caffeine_type"     "procrastination"   "multitasking"

Descriptives

surveydata |> 
  summarise(
    mheight = mean(height, na.rm = T),
    sdheight = sd(height, na.rm = T),
    n = sum(!is.na(height))
  )

# A tibble: 1 × 3
  mheight sdheight     n
    <dbl>    <dbl> <int>
1    168.     8.27   537

Assumptions

The shapiro.test() suggests that our assumption of normality is not okay!
(the p-value is $<.05$, suggesting that we reject the hypothesis that the data are drawn from a normally distributed population)

shapiro.test(surveydata$height)


    Shapiro-Wilk normality test

data:  surveydata$height
W = 0.99063, p-value = 0.001731

However, as always, visualisations are vital here. The histogram below doesn’t look all that great, but the t.test is quite robust against slight violations of normality, especially as sample sizes increase beyond 30, and our data here actually looks fairly normal (this is a judgement call here - over time you will start to get a sense of what you might deem worrisome in these plots!).

ggplot(data = surveydata, aes(x = height)) + 
  geom_histogram() +
  # adding our hypothesised mean
  geom_vline(xintercept = 165)

We can also take a quick look at the QQplot. The points follow the line closely apart from at the tail ends, matching the heavier tails of the distribution that are visible in the histogram above.

qqnorm(surveydata$height)
qqline(surveydata$height)

The data are not very skewed, and together with the fact that we are working with a sample of 537, i feel fairly satisfied that the $t$-test will lead us to valid inferences.

t-test

t.test(surveydata$height, mu = 165, alternative = "two.sided")


    One Sample t-test

data:  surveydata$height
t = 7.7102, df = 536, p-value = 6.146e-14
alternative hypothesis: true mean is not equal to 165
95 percent confidence interval:
 167.0508 168.4531
sample estimates:
mean of x 
 167.7519

Names and Tips

Research Question Can a server earn higher tips simply by introducing themselves by name when greeting customers?

Researchers investigated the effect of a server introducing herself by name on restaurant tipping. The study involved forty, 2-person parties eating a $23.21 fixed-price buffet Sunday brunch at Charley Brown’s Restaurant in Huntington Beach, California, on April 10 and 17, 1988.

Each two-person party was randomly assigned by the waitress to either a name or a no-name condition. The total amount paid by each party at the end of their meal was then recorded.

The data are available at https://uoepsy.github.io/data/gerritysim.csv. (This is a simulated example based on Garrity and Degelman (1990))

Question 11

Conduct an independent samples $t$-test to assess whether higher tips were earned when the server introduced themselves by name, in comparison to when they did not.

Hints

There is a direction in the research question stated above, which means we will want to set alternative = ??.
We’ll want to check the normality (either visually or with a test) of the variable of interest for each group.
Some researchers suggest using the Welch t-test by default. This means you can relax the assumption of equal variances in the groups. If you want to test whether two variances are equal, try the var.test() function.

Data

tipdata <- read_csv("https://uoepsy.github.io/data/gerritysim.csv")
dim(tipdata)

[1] 40  3

head(tipdata)

# A tibble: 6 × 3
   paid condition party
  <dbl> <chr>     <dbl>
1  28.5 name          1
2  27   no name       2
3  24.1 no name       3
4  28   name          4
5  30   name          5
6  27.5 name          6

It might be nice to conduct our analysis on just the tip given, and not the $23.21 meal price + tip.

#make a "tip" column, which is minus the meal amount
tipdata <- 
  tipdata |> mutate(
    tip = paid - 23.21
  )

Descriptives

tipdata |> 
  group_by(condition) |>
  summarise(
    meantip = mean(tip),
    sdtip = sd(tip)
  )

# A tibble: 2 × 3
  condition meantip sdtip
  <chr>       <dbl> <dbl>
1 name         4.94  1.88
2 no name      3.18  1.35

Plot

ggplot(data = tipdata, aes(x = tip, y = condition)) +
  geom_boxplot()

Assumptions

According to these tests, we have normally distributed data for both groups, with equal variances.

shapiro.test(tipdata$tip[tipdata$condition=="name"])


    Shapiro-Wilk normality test

data:  tipdata$tip[tipdata$condition == "name"]
W = 0.96267, p-value = 0.5985

shapiro.test(tipdata$tip[tipdata$condition=="no name"])


    Shapiro-Wilk normality test

data:  tipdata$tip[tipdata$condition == "no name"]
W = 0.94405, p-value = 0.2857

var.test(tip ~ condition, data = tipdata)


    F test to compare two variances

data:  tip by condition
F = 1.9344, num df = 19, denom df = 19, p-value = 0.1595
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.7656473 4.8870918
sample estimates:
ratio of variances 
           1.93437

t-test

Because the variances do not appear to be unequal, we can actually use the standard t-test with var.equal = TRUE if we want. However, we’ll continue with the Welch t-test.

Remember that our alternative hypothesis here is that the average tips in the “name” condition is greater than in the “no name” condition.
R will take the levels in order here (alphabetically), and assume that the alternative is for that group, so we use alternative = "greater" here to say that the alternative is $\text{name}-\text{no name} > 0$.

t.test(tip ~ condition, data = tipdata, alternative = "greater")


    Welch Two Sample t-test

data:  tip by condition
t = 3.4117, df = 34.502, p-value = 0.0008314
alternative hypothesis: true difference in means between group name and group no name is greater than 0
95 percent confidence interval:
 0.8893105       Inf
sample estimates:
   mean in group name mean in group no name 
               4.9450                3.1825

Optional Extras

Here are a few extra questions for you to practice performing tests and making plots:

Optional Extra 1

Are dogs heavier on average than cats?
Data from Week 1: https://uoepsy.github.io/data/pets_seattle.csv

Hints

Remember from week 1 - not everything in that data is either a cat or a dog!

Optional Extra 2

Is taking part in a cognitive behavioural therapy (CBT) based programme associated with a greater reduction, on average, in anxiety scores in comparison to a Control group?
Data are at https://uoepsy.github.io/data/cbtanx.csv. The dataset contains information on each person in an organisation, recording their professional role (management vs employee), whether they are allocated into the CBT programme or not (control vs cbt), and scores on anxiety at both the start and the end of the study period.

Hints

you might have to make a new variable in order to test the research question.

Optional Extra 3

Are students on our postgraduate courses shorter/taller than those on our undergraduate courses?
We can again use the data from the past surveys: https://uoepsy.github.io/data/surveydata_historical.csv.

“USMR” is our only postgraduate course.

Hints

You’ll need to create a variable that identifies whether a respondent is from a postgrad or and undergrad course.

Exercises: T-tests

Intervals again

Procrastination Scores

Data

Descriptives

Assumptions

Heights

Data

Descriptives

Assumptions

t-test

Names and Tips

Data

Descriptives

Plot

Assumptions

t-test

Optional Extras

Footnotes