The answers will be made available for discussing particular questions right after the mock.
Load the data into R before starting the mock or exam. Follow the data-loading instructions on NOW.
For the exam, a NOW News message one hour before the start will tell you which datasets to load. Have these loaded before the exam begins.
Use an R script for your attempt. Keep it available to compare with the answer script afterwards.
Leave a few minutes to check your answers and submit within your time allowance.
If you finish early, check your answers and identify one result you can’t explain in words.
Take a 10-minute break after the mock.
Keep your attempt available so you can compare it with the worked answers.
Download week_09_workshop_answer.R when it is released on NOW and move it into your RStudio Project folder.
We will discuss particular problems you raise. The question-by-question worked slides are available for these discussions and later revision.
Find the study guide Statistical concepts, tests and R in NOW → Assessments. It brings together the module topics, key concepts, tests, R commands and relevant formulas.
Choose a question you found difficult. Decide what you need to practise:
Use the guide to revisit that topic, practise the matching workshop or problem-set task, then retry without the answers.
d_1 <- read_csv("workshop_data/assessment_dataset_01.csv")
m_1 <- t.test(value ~ section, data = d_1,
var.equal = TRUE, conf.level = .80)
m_1$statistic
# t # -0.4489152
Answer: t = -0.4489. The difference is 1st minus 2nd; the negative sign means the first sample mean is lower.
The null hypothesis is a population mean difference of zero. Equal variances are specified in the question.
Revisit: Weeks 3–4 workshop and lecture slides — equal-variance t-tests and reading the output.
m_1$parameter
# df # 998
Answer: df = 998, the two sample sizes added together minus 2.
The degrees of freedom determine the shape of the t-distribution used for this test.
Revisit: Weeks 3–4 workshop and lecture slides — t-test degrees of freedom.
m_1$conf.int
# [1] -2.397991 1.154434 # attr(,"conf.level") # [1] 0.8
Answer: [-2.398, 1.154], for population mean 1st minus 2nd.
Zero is inside the interval: these data do not rule out no population difference at this confidence level.
Use conf.level = .80; the default 95% interval answers a different question.
Revisit: Week 4 workshop and lecture slides — confidence intervals and confidence levels.
# Q4: area ABOVE 1.75 pt(1.75, df = 14, lower.tail = FALSE)
# [1] 0.05099653
# Q5: area BELOW 0.75 pt(0.75, df = 5)
# [1] 0.7564879
Q4: .051. Q5: .756 (values shown here to three decimal places).
These ask for one-tail probabilities, not two-sided p-values. Use the supplied t-value and df for each question.
Revisit: Week 3 workshop and lecture slides — t-distribution tails and pt().
Two independent samples; assume equal population variances.
n_x <- 12; n_y <- 10
avg_x <- 100.589; avg_y <- 101.356
sd_x <- 1.297; sd_y <- 1.611
df <- n_x + n_y - 2
pooled_variance <- ((n_x - 1) * sd_x^2 +
(n_y - 1) * sd_y^2) / df
se <- sqrt(pooled_variance) * sqrt(1 / n_x + 1 / n_y)
avg_diff <- avg_x - avg_y
Pool the sample variances to estimate the common population variance. Keep the requested difference x minus y throughout.
Revisit: Week 4 workshop and lecture slides — pooled variance, standard error and the t-statistic.
avg_diff / se
# [1] -1.238165
Answer: t = -1.2382. The sample mean difference is -0.767, about 1.24 standard errors below the null difference of zero.
The negative sign comes from x minus y. Reversing the subtraction reverses the sign.
Revisit: Week 4 workshop and lecture slides — pooled variance, standard error and the t-statistic.
tau <- qt(.95, df = df) avg_diff - tau * se
# [1] -1.835402
avg_diff + tau * se
# [1] 0.3014024
Answer: [-1.835, 0.301], for the population difference mu_x minus mu_y.
A central 90% interval leaves 5% in each tail. Use the cutoff with .95 below it, not .90.
Revisit: Week 4 workshop and lecture slides — confidence intervals and qt().
d_3 <- read_csv("workshop_data/assessment_dataset_03.csv")
m_2 <- lm(y ~ x + z, data = d_3)
m_2_summary <- summary(m_2)
m_2_summary$fstatistic
# value numdf dendf # 2.859026 2.000000 47.000000
Answer: F = 2.859, with df 2 and 47. Read value; the other two numbers are the degrees of freedom.
The null hypothesis is that both predictor coefficients are zero, comparing this model with an intercept-only model.
Revisit: Week 6 workshop and lecture slides — summary() and the overall F-test.
new_values <- tibble(x = 10, z = 12) new_values
# # A tibble: 1 × 2 # x z # <dbl> <dbl> # 1 10 12
predict(m_2, newdata = new_values)
# 1 # 17.15589
Answer: 17.1559 (about 17.2).
Supply both predictors, using their original names. This is the predicted mean of y at those predictor values.
Revisit: Week 5 workshop and lecture slides — multiple regression and predict().
sigma(m_2)
# [1] 52.62584
Answer: 52.6258 (about 52.63).
This is the estimated residual standard deviation: spread around the predicted mean, in y’s units.
The model assumes the same residual SD at every x and z. It is different from the predicted mean and from uncertainty about that mean.
Revisit: Week 5 workshop and lecture slides — sigma and residual spread.
These supplied numbers describe a separate model with one numeric predictor and an intercept.
n <- 25; K <- 1 rss <- 4186.153; tss <- 19560.807 ess <- tss - rss rsq <- ess / tss rsq
# [1] 0.7859928
Answer: R-squared = 0.786: about 78.6% explained and 21.4% unexplained. TSS = ESS + RSS.
Revisit: Week 6 workshop and lecture slides — ESS, RSS, TSS and R-squared.
Df1 <- K Df2 <- n - K - 1 (ess / Df1) / (rss / Df2)
# [1] 84.47303
Answer: F = 84.47, with df 1 and 23.
Compare explained variation per predictor df with unexplained variation per residual df.
Here the simpler model is intercept-only: its RSS equals TSS. The RSS reduction is ESS, linking this calculation to the model comparisons in Week 8.
Revisit: Week 6 workshop and lecture slides — overall F and its degrees of freedom.
1 - (1 - rsq) * (n - 1) / (n - K - 1)
# [1] 0.7766882
Answer: adjusted R-squared = 0.777.
It is below the unadjusted value of 0.786 because the adjustment penalises model complexity relative to the available sample size.
Use n = 25 and K = 1 supplied for this question, rather than values from the earlier fitted model.
Revisit: Week 6 workshop and lecture slides — adjusted R-squared.
d_ano <- read_csv("workshop_data/anorexia2.csv")
m_3 <- lm(Postwt ~ Treat, data = d_ano)
coef(m_3)
# (Intercept) Treatctrl Treatfam # 85.696552 -4.588859 4.797566
The reference group is cbt, so its predicted mean equals the intercept.
Treatfam combines the predictor name Treat and level fam; it gives the difference from cbt.
Revisit: Week 7 workshop and lecture slides — categorical predictors and group means.
predict(m_3, newdata = tibble(Treat = "cbt"))
# 1 # 85.69655
Answer: 85.6966.
This is the predicted group mean, rather than a difference between groups.
Revisit: Week 7 workshop and lecture slides — categorical predictors and group means.
mean_fam <- predict(m_3, newdata = tibble(Treat = "fam")) mean_cbt <- predict(m_3, newdata = tibble(Treat = "cbt")) mean_fam - mean_cbt
# 1 # 4.797566
Answer: 4.7976. The predicted mean for fam is about 4.80 units higher than for cbt.
This also equals the Treatfam coefficient. It is a difference, rather than the predicted mean of fam.
Revisit: Week 7 workshop and lecture slides — group differences and reference levels.
m_4 <- lm(Postwt ~ Treat + Prewt, data = d_ano) coef(m_4)
# (Intercept) Treatctrl Treatfam Prewt # 49.7711090 -4.0970655 4.5630627 0.4344612
The model has group differences plus a common Prewt slope. The question does not request an interaction.
For fam: predicted Postwt = intercept + fam difference + Prewt slope × Prewt.
Revisit: Week 8 workshop and lecture slides — categorical and continuous predictors together.
new_person <- tibble(Treat = "fam", Prewt = 80) new_person
# # A tibble: 1 × 2 # Treat Prewt # <chr> <dbl> # 1 fam 80
predict(m_4, newdata = new_person)
# 1 # 89.09106
Answer: 89.0911.
Using the printed coefficients: 49.7711 + 4.5631 + 0.4344612 × 80 ≈ 89.0911.
Revisit: Week 8 workshop and lecture slides — categorical and continuous predictors together.