EPI 553 — Tests of Hypotheses Lab Due: End of class
This lab uses a Driver-Navigator pair programming approach. You will work with a partner, taking turns in two roles:
Round 1 (Tasks 1–3): Student A is the Driver, Student B is the Navigator. Round 2 (Tasks 4–6): Student B is the Driver, Student A is the Navigator.
Switch roles after completing Task 3. Both partners should understand all tasks — the Navigator should actively participate by checking the work, asking questions, and discussing interpretations.
Submission: Each pair submits one knitted HTML file with both partners’ names. Upload to Brightspace by end of class.
Use the same BRFSS 2020 dataset from the guided practice.
# Load packages and data
library(tidyverse)## Warning: package 'tidyverse' was built under R version 4.5.2## Warning: package 'ggplot2' was built under R version 4.5.2## Warning: package 'tibble' was built under R version 4.5.2## Warning: package 'tidyr' was built under R version 4.5.2## Warning: package 'readr' was built under R version 4.5.2## Warning: package 'purrr' was built under R version 4.5.2## Warning: package 'dplyr' was built under R version 4.5.2## Warning: package 'stringr' was built under R version 4.5.2## Warning: package 'forcats' was built under R version 4.5.2## Warning: package 'lubridate' was built under R version 4.5.2## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
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## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(broom)## Warning: package 'broom' was built under R version 4.5.2library(knitr)## Warning: package 'knitr' was built under R version 4.5.2library(kableExtra)## Warning: package 'kableExtra' was built under R version 4.5.2##
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## somelibrary(gtsummary)## Warning: package 'gtsummary' was built under R version 4.5.2brfss_mlr <- readRDS(
"C:/Users/abbym/OneDrive/Desktop/STATS553/R Materials/epi553/scripts/brfss_mlr_2020.rds"
)| Task | Description | Points |
|---|---|---|
| Task 1 | Fitting Models and ANOVA Tables | 15 |
| Task 2 | Type I vs. Type III Sums of Squares | 15 |
| Task 3 | Partial F-Tests for Individual Variables | 20 |
| Switch roles | ||
| Task 4 | T-Tests and the F-Test Equivalence | 15 |
| Task 5 | Chunk Test (Testing Groups of Variables) | 20 |
| Task 6 | Synthesis and Interpretation | 15 |
| Total | 100 |
Use tidy() with conf.int = TRUE to display
the coefficients. Report the fitted equation with rounded
coefficients.
model_a <- lm(menthlth_days ~ physhlth_days + sleep_hrs + age,
data = brfss_mlr)
tidy(model_a, conf.int = TRUE) %>%
mutate(across(where(is.numeric), ~ round(., 4))) %>%
kable(
caption = "Table 1. Model A Coefficients",
col.names = c("Term", "Estimate", "Std. Error", "t-statistic", "p-value", "95% CI Lower", "95% CI Upper")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)| Term | Estimate | Std. Error | t-statistic | p-value | 95% CI Lower | 95% CI Upper |
|---|---|---|---|---|---|---|
| (Intercept) | 10.6684 | 0.6102 | 17.4844 | 0 | 9.4722 | 11.8646 |
| physhlth_days | 0.3182 | 0.0131 | 24.2179 | 0 | 0.2924 | 0.3439 |
| sleep_hrs | -0.5063 | 0.0760 | -6.6630 | 0 | -0.6553 | -0.3573 |
| age | -0.0800 | 0.0059 | -13.4532 | 0 | -0.0916 | -0.0683 |
# Calculate Type I SS
anova_type1 <- anova(model_a)
anova_type1 %>%
as.data.frame() %>%
rownames_to_column("Source") %>%
mutate(across(where(is.numeric), ~ round(., 2))) %>%
kable(
caption = "Table 2. Type I (Sequential) Sums of Squares — anova(), Model A",
col.names = c("Source", "df", "Sum of Sq", "Mean Sq", "F value", "p-value")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
footnote(general = "Variables are tested in the order they appear in the model formula.")| Source | df | Sum of Sq | Mean Sq | F value | p-value |
|---|---|---|---|---|---|
| physhlth_days | 1 | 29474.75 | 29474.75 | 576.00 | 0 |
| sleep_hrs | 1 | 3323.04 | 3323.04 | 64.94 | 0 |
| age | 1 | 9261.47 | 9261.47 | 180.99 | 0 |
| Residuals | 4996 | 255652.08 | 51.17 | NA | NA |
| Note: | |||||
| Variables are tested in the order they appear in the model formula. |
# Calculate model SS
type1_ss <- anova_type1$`Sum Sq`
ssr_full <- sum(type1_ss[1:3]) # Model SS (sum of all predictor Type I SS)
sse_full <- type1_ss[4] # Residual SS
cat("SSR (Model):", round(ssr_full, 2), "\n")## SSR (Model): 42059.26# Calculate Type III SS
# Type III using car::Anova()
anova_type3 <- Anova(model_a, type = "III")
# Side-by-side comparison
comparison <- tibble(
Variable = c("physhlth_days", "sleep_hrs", "age"),
`Type I SS` = round(anova_type1$`Sum Sq`[1:3], 1),
`Type III SS` = round(anova_type3$`Sum Sq`[2:4], 1)
)
comparison %>%
kable(caption = "Table 3. Type I vs. Type III Sums of Squares, Model A") %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
footnote(general = "Type I SS depends on variable entry order; Type III SS does not. The last variable (age) has identical values.")| Variable | Type I SS | Type III SS |
|---|---|---|
| physhlth_days | 29474.7 | 30012.3 |
| sleep_hrs | 3323.0 | 2271.8 |
| age | 9261.5 | 9261.5 |
| Note: | ||
| Type I SS depends on variable entry order; Type III SS does not. The last variable (age) has identical values. |
1a. (5 pts) Fit the following model:
\[\text{menthlth\_days} = \beta_0 +
\beta_1 \cdot \text{physhlth\_days} + \beta_2 \cdot \text{sleep\_hrs} +
\beta_3 \cdot \text{age} + \varepsilon\] \[\text{menthlth\_days} = 10.6684 +
0.3182\text{physhlth\_days} + -0.5063 \text{sleep\_hrs} +
-0.0800 \text{age} + \varepsilon\] 1b. (5
pts) Use anova() on this model to obtain the Type
I (sequential) sums of squares. Present the ANOVA table and verify that
the sum of all predictor Type I SS equals the model SSR. Show this
calculation explicitly. 29474.75 + 3323.04 + 9261.47= 42,059.26
1c. (5 pts) Use car::Anova() with
type = "III" on the same model to obtain the Type III sums
of squares. Compare the Type I and Type III SS for each variable. Which
variable’s SS is the same in both tables? Why? Age is the same in both
tables, because type I is based on variables that come after the
variable, and since age is the last variable in the model, it is the
same for both type I and type III. —
# Model B
model_b <- lm(menthlth_days ~ age + sleep_hrs + physhlth_days,
data = brfss_mlr)
# Fitted Model
tidy(model_b, conf.int = TRUE) %>%
mutate(across(where(is.numeric), ~ round(., 4))) %>%
kable(
caption = "Table 4. Model B Coefficients",
col.names = c("Term", "Estimate", "Std. Error", "t-statistic", "p-value", "95% CI Lower", "95% CI Upper")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)| Term | Estimate | Std. Error | t-statistic | p-value | 95% CI Lower | 95% CI Upper |
|---|---|---|---|---|---|---|
| (Intercept) | 10.6684 | 0.6102 | 17.4844 | 0 | 9.4722 | 11.8646 |
| age | -0.0800 | 0.0059 | -13.4532 | 0 | -0.0916 | -0.0683 |
| sleep_hrs | -0.5063 | 0.0760 | -6.6630 | 0 | -0.6553 | -0.3573 |
| physhlth_days | 0.3182 | 0.0131 | 24.2179 | 0 | 0.2924 | 0.3439 |
# Type I
anova_type1 <- anova(model_b)
anova_type1 %>%
as.data.frame() %>%
rownames_to_column("Source") %>%
mutate(across(where(is.numeric), ~ round(., 2))) %>%
kable(
caption = "Table 5. Type I (Sequential) Sums of Squares — anova(), Model B",
col.names = c("Source", "df", "Sum of Sq", "Mean Sq", "F value", "p-value")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
footnote(general = "Variables are tested in the order they appear in the model formula.")| Source | df | Sum of Sq | Mean Sq | F value | p-value |
|---|---|---|---|---|---|
| age | 1 | 7249.10 | 7249.10 | 141.66 | 0 |
| sleep_hrs | 1 | 4797.91 | 4797.91 | 93.76 | 0 |
| physhlth_days | 1 | 30012.26 | 30012.26 | 586.51 | 0 |
| Residuals | 4996 | 255652.08 | 51.17 | NA | NA |
| Note: | |||||
| Variables are tested in the order they appear in the model formula. |
car::Anova(model_b, type = "III")## Anova Table (Type III tests)
##
## Response: menthlth_days
## Sum Sq Df F value Pr(>F)
## (Intercept) 15643 1 305.706 < 2.2e-16 ***
## age 9261 1 180.989 < 2.2e-16 ***
## sleep_hrs 2272 1 44.396 2.972e-11 ***
## physhlth_days 30012 1 586.505 < 2.2e-16 ***
## Residuals 255652 4996
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# Calculate Type III SS
# Type III using car::Anova()
anova_type3 <- Anova(model_b, type = "III")
# Side-by-side comparison
comparison <- tibble(
Variable = c("age", "sleep_hrs", "physhlth_days"),
`Type I SS` = round(anova_type1$`Sum Sq`[1:3], 1),
`Type III SS` = round(anova_type3$`Sum Sq`[2:4], 1)
)
comparison %>%
kable(caption = "Table 6. Type I vs. Type III Sums of Squares, Model B") %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE) %>%
footnote(general = "Type I SS depends on variable entry order; Type III SS does not. The last variable (physhlth_days) has identical values.")| Variable | Type I SS | Type III SS |
|---|---|---|
| age | 7249.1 | 9261.5 |
| sleep_hrs | 4797.9 | 2271.8 |
| physhlth_days | 30012.3 | 30012.3 |
| Note: | ||
| Type I SS depends on variable entry order; Type III SS does not. The last variable (physhlth_days) has identical values. |
2a. (5 pts) Fit the same model from Task 1 but reverse the variable order:
\[\text{menthlth\_days} = \beta_0 +
\beta_1 \cdot \text{age} + \beta_2 \cdot \text{sleep\_hrs} + \beta_3
\cdot \text{physhlth\_days} + \varepsilon\] \[\text{menthlth\_days} = 10.6684 +
-0.0800 \text{age} + -0.5063 \text{sleep\_hrs} +
0.3182 \text{physhlth\_days} + \varepsilon\] Run
anova() on this model and compare the Type I SS to what you
obtained in Task 1b. Which values changed and which stayed the same?
None of the values stayed the same. 2b. (5 pts) Run
car::Anova(type = "III") on this reordered model. Did the
Type III SS change compared to Task 1c? Explain why or why not. The type
III SS did not change compared to task 1c, because the order of the
variables does not matter when evaluating type III error. 2c. (5
pts) In 2–3 sentences, explain when an epidemiologist should
prefer Type III SS over Type I SS. Give a concrete example from public
health research where the choice matters. An epidemiologist should
prefer Type III over Type I SS when they are wondering what an
additional variable will do to a whole model, rather than its specific
additional impact it has on the model. For example, an epidemiologist
might want to use type III over type I when they are looking to see the
whole impact of a variable on the model. For example, they may want to
use type III for something like seeing if age will impact calcium intake
and bone mineral density, as it shouldn’t matter the order it goes in.
If it were something that required more of a sequential order, like the
impact of smoking, then age, on lung cancer outcome, they may want to
use Type I.
—
# Partial F-test
model_full <- lm(menthlth_days ~ physhlth_days + sleep_hrs + age,
data = brfss_mlr)
model_reduced <- lm(menthlth_days ~ physhlth_days + sleep_hrs,
data = brfss_mlr)
anova(model_reduced, model_full)## Analysis of Variance Table
##
## Model 1: menthlth_days ~ physhlth_days + sleep_hrs
## Model 2: menthlth_days ~ physhlth_days + sleep_hrs + age
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 4997 264914
## 2 4996 255652 1 9261.5 180.99 < 2.2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# Manual Calculation
ssr_full <- sum(anova(model_full)$`Sum Sq`[1:3])
ssr_reduced <- sum(anova(model_reduced)$`Sum Sq`[1:2])
mse_full <- anova(model_full)$`Mean Sq`[4] # MSE of full model
F_stat <- (ssr_full - ssr_reduced) / 1 / mse_full
cat("SSR(full):", round(ssr_full, 2), "\n")## SSR(full): 42059.26cat("SSR(reduced):", round(ssr_reduced, 2), "\n")## SSR(reduced): 32797.79cat("SS(Age | others):", round(ssr_full - ssr_reduced, 2), "\n")## SS(Age | others): 9261.47cat("MSE(full):", round(mse_full, 2), "\n")## MSE(full): 51.17cat("F-statistic:", round(F_stat, 4), "\n")## F-statistic: 180.9894cat("Critical value F(1, 4996, 0.95):", round(qf(0.95, 1, 4996), 4), "\n")## Critical value F(1, 4996, 0.95): 3.8433cat("p-value:", format.pval(pf(F_stat, 1, 4996, lower.tail = FALSE)), "\n")## p-value: < 2.22e-163a. (10 pts) Conduct a partial F-test to determine
whether age adds significantly to the prediction of mental
health days, given that physhlth_days and
sleep_hrs are already in the model. Do this by:
age)age)anova(reduced, full) to compare themState the null hypothesis, report the F-statistic and p-value, and
state your conclusion at \(\alpha =
0.05\). The null hypothesis is that age does not add
statistically significant information to the model. The F-statistic is
180.99 and the p-value is < 2.2e-16, which means you reject the null
hypothesis, meaning age does add statistically significant information
to the model. 3b. (10 pts) Now verify your result from
3a manually. Using the anova() output from the full model
(Task 1b), identify \(SS(\text{age} \mid
\text{physhlth\_days}, \text{sleep\_hrs})\) from the Type I
table. Compute the F-statistic as:
\[F = \frac{SS(\text{age} \mid
\text{physhlth\_days}, \text{sleep\_hrs}) / 1}{MSE(\text{full
model})}\] F= [(29474.75+3323.04+9261.47)/1]/ Compare to the
critical value \(F_{1, n-p-1, 0.95}\).
Does your manual calculation agree with the anova()
comparison from 3a? Yes, it does agree with the anova comparison from
3a. They are both 180.99. —
⟳ Switch roles now!
# Get t-statistics from summary()
t_stats <- tidy(model_full) %>%
filter(term != "(Intercept)") %>%
select(term, t_stat = statistic, t_pvalue = p.value)
# Get F-statistics from Type III
f_stats <- Anova(model_full, type = "III") %>%
as.data.frame() %>%
rownames_to_column("term") %>%
filter(!term %in% c("(Intercept)", "Residuals")) %>%
select(term, f_stat = `F value`, f_pvalue = `Pr(>F)`)
# Compare
left_join(t_stats, f_stats, by = "term") %>%
mutate(
`t²` = round(t_stat^2, 4),
f_stat = round(f_stat, 4),
`t² = F?` = ifelse(abs(t_stat^2 - f_stat) < 0.001, "✓", "✗"),
t_pvalue = round(t_pvalue, 6),
f_pvalue = round(f_pvalue, 6),
`p-values equal?` = ifelse(abs(t_pvalue - f_pvalue) < 0.0001, "✓", "✗")
) %>%
select(term, t_stat = t_stat, `t²`, `F (Type III)` = f_stat, `t² = F?`,
`p (t-test)` = t_pvalue, `p (F-test)` = f_pvalue, `p-values equal?`) %>%
mutate(t_stat = round(t_stat, 4)) %>%
kable(caption = "Table 7. Equivalence of T-Tests and Type III Partial F-Tests") %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)| term | t_stat | t² | F (Type III) | t² = F? | p (t-test) | p (F-test) | p-values equal? |
|---|---|---|---|---|---|---|---|
| physhlth_days | 24.2179 | 586.5051 | 586.5051 | ✓ | 0 | 0 | ✓ |
| sleep_hrs | -6.6630 | 44.3961 | 44.3961 | ✓ | 0 | 0 | ✓ |
| age | -13.4532 | 180.9894 | 180.9894 | ✓ | 0 | 0 | ✓ |
summary(model_full)##
## Call:
## lm(formula = menthlth_days ~ physhlth_days + sleep_hrs + age,
## data = brfss_mlr)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.1227 -3.4457 -1.8464 0.0493 30.3922
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.668370 0.610164 17.484 < 2e-16 ***
## physhlth_days 0.318159 0.013137 24.218 < 2e-16 ***
## sleep_hrs -0.506317 0.075989 -6.663 2.97e-11 ***
## age -0.079966 0.005944 -13.453 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.153 on 4996 degrees of freedom
## Multiple R-squared: 0.1413, Adjusted R-squared: 0.1408
## F-statistic: 274 on 3 and 4996 DF, p-value: < 2.2e-164a. (5 pts) Using the full model from Task 1
(menthlth_days ~ physhlth_days + sleep_hrs + age), run
summary() and extract the t-statistics and p-values for
each coefficient.
4b. (5 pts) For each predictor, compute \(t^2\) and compare it to the Type III F-statistic from Task 1c. Create a table showing the t-statistic, \(t^2\), the Type III F-statistic, and both p-values. Are they equivalent? Yes, they are equivalent. 4c. (5 pts) In your own words, explain why the t-test and the Type III partial F-test give the same result. What is the fundamental relationship between the t-distribution and the F-distribution that makes this true? The t-test and F-test give the same result because the null hypotheses are being tested while considering all other variables. Neither of them are order-dependent, either. They are testing the same things, while using the same statistical contributions. —
# Test group
# Reduced model:
model_partial <- lm(menthlth_days ~ physhlth_days + sleep_hrs + age, data = brfss_mlr)
# Full model: health behaviors + demographics
m_full <- lm(menthlth_days ~ physhlth_days + sleep_hrs + age + income_cat + sex + exercise,
data = brfss_mlr)
tidy(m_full, conf.int = TRUE) %>%
mutate(across(where(is.numeric), ~ round(., 4))) %>%
kable(
caption = "Table 1. Model A Coefficients",
col.names = c("Term", "Estimate", "Std. Error", "t-statistic", "p-value", "95% CI Lower", "95% CI Upper")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)| Term | Estimate | Std. Error | t-statistic | p-value | 95% CI Lower | 95% CI Upper |
|---|---|---|---|---|---|---|
| (Intercept) | 12.4755 | 0.7170 | 17.4006 | 0.0000 | 11.0699 | 13.8810 |
| physhlth_days | 0.2917 | 0.0136 | 21.4779 | 0.0000 | 0.2650 | 0.3183 |
| sleep_hrs | -0.5092 | 0.0753 | -6.7574 | 0.0000 | -0.6569 | -0.3614 |
| age | -0.0823 | 0.0059 | -13.8724 | 0.0000 | -0.0939 | -0.0707 |
| income_cat | -0.3213 | 0.0520 | -6.1778 | 0.0000 | -0.4233 | -0.2194 |
| sexFemale | 1.2451 | 0.2023 | 6.1535 | 0.0000 | 0.8484 | 1.6417 |
| exerciseYes | -0.3427 | 0.2531 | -1.3537 | 0.1759 | -0.8389 | 0.1536 |
# Chunk test
anova(model_partial, m_full) %>%
as.data.frame() %>%
rownames_to_column("Model") %>%
mutate(
Model = c("Reduced (physhlth + sleep + age only)", "Full (+ demographics)"),
across(where(is.numeric), ~ round(., 4))
) %>%
kable(
caption = "Table 8. Chunk Test: Do demographic variables collectively add to the model?",
col.names = c("Model", "Res. df", "RSS", "df", "Sum of Sq", "F", "p-value")
) %>%
kable_styling(bootstrap_options = c("striped", "hover"), full_width = FALSE)| Model | Res. df | RSS | df | Sum of Sq | F | p-value |
|---|---|---|---|---|---|---|
| Reduced (physhlth + sleep + age only) | 4996 | 255652.1 | NA | NA | NA | NA |
| Full (+ demographics) | 4993 | 250992.8 | 3 | 4659.276 | 30.8957 | 0 |
# Manual Computation
ssr_full <- sum(anova(m_full)$`Sum Sq`[1:6])
ssr_reduced <- sum(anova(model_partial)$`Sum Sq`[1:3])
mse_full <- anova(m_full)$`Mean Sq`[7]
df_diff <- 4 # 4 additional variables
F_chunk <- ((ssr_full - ssr_reduced) / df_diff) / mse_full
cat("SSR(full):", round(ssr_full, 2), "\n")## SSR(full): 46718.54cat("SSR(reduced):", round(ssr_reduced, 2), "\n")## SSR(reduced): 42059.26cat("Difference:", round(ssr_full - ssr_reduced, 2), "\n")## Difference: 4659.28cat("df (number of added variables):", df_diff, "\n")## df (number of added variables): 4cat("MSE(full):", round(mse_full, 2), "\n")## MSE(full): 50.27cat("F-statistic:", round(F_chunk, 4), "\n")## F-statistic: 23.1717cat("Critical value F(4, 4993, 0.95):", round(qf(0.95, df_diff, 4993), 4), "\n")## Critical value F(4, 4993, 0.95): 2.3737cat("p-value:", format.pval(pf(F_chunk, df_diff, 4993, lower.tail = FALSE)), "\n")## p-value: < 2.22e-165a. (10 pts) Now consider the full 6-predictor model:
\[\text{menthlth\_days} = \beta_0 + \beta_1 \cdot \text{physhlth\_days} + \beta_2 \cdot \text{sleep\_hrs} + \beta_3 \cdot \text{age} + \beta_4 \cdot \text{income\_cat} + \beta_5 \cdot \text{sex} + \beta_6 \cdot \text{exercise} + \varepsilon\]
Test whether income_cat, sex, and
exercise — as a group — significantly add to the prediction
of mental health days, given that physhlth_days,
sleep_hrs, and age are already in the
model.
State the null hypothesis (in both words and mathematical notation), conduct the test, and state your conclusion. Null Hypothesis:H0:βincome=βsex=βexercise=0; The group adds nothing statistically significant to the model. Alternative Hypothesis: H1:βincome=βsex=βexercise does not equal 0; The group improves the prediction ability of the model. 5b. (5 pts) Compute the chunk test F-statistic manually using:
\[F = \frac{\{SSR(\text{full}) - SSR(\text{reduced})\} / (df_{\text{full}} - df_{\text{reduced}})}{MSE(\text{full})}\]
Show all intermediate values. Does your manual computation match the
anova() result?
5c. (5 pts) Note that exercise was
not individually significant in the Type III table, yet
it is part of a group that is collectively significant. In 2–3
sentences, explain how this is possible and what it means for model
building in epidemiology. It is possible for something to not be
statistically significant individually, but as a part of a group it is.
This is because combined with the other variables, it can help to
explain the model, but alone, it doesn’t contribute much. This means
that even though something may not be individually important, it should
be looked at as a group with other variables when building models in
epidemiology. —
6a. (5 pts) Based on the full model, which
predictors are statistically significant at \(\alpha = 0.05\)? List them and briefly
state the direction of each association (positive or negative). The
predictors that are statistically significant are physical health days,
which have a positive association, hours of sleep, which has a negative
association, age , which has a negative association, income category
which has a negative association, and sex, which has a positive
association with being female. 6b. (5 pts) A colleague
argues: “We should drop exercise from the model because
it’s not significant.” Do you agree? Write a 2–3 sentence response
explaining your reasoning. Consider the chunk test results and
epidemiologic rationale. I don’t agree, because it was shown that will
the group, it did have an impact on the model and helped to make the
whole model a better predictor of bad mental health days. Even though it
is not individually significant, it still plays a role in increasing the
understanding of bad mental health days in the entire model. 6c.
(5 pts) Write a 3–4 sentence summary of the hypothesis testing
results for a non-statistical audience (e.g., a public health program
manager). Your summary should convey which factors were identified as
independently associated with mental health days and which were not,
without using jargon like “p-value,” “F-test,” or “sums of squares.” The
number of bad physical health days, hours of sleep, age, income
category, and sex were determined to be the most important factors to
consider when predicting how many bad mental health days a person will
have in a month. Exercise was determined to not be as important when
predicting how many bad mental health days a person will have in a
month. This was shown through a series of tests that determined if the
combination of variables/individual variables were helpful in creating a
model that could predict the number of bad mental health days a person
will have in a month. ### Task 7: Reflection (Not Graded, strongly
recommended) In 2–3 sentences, reflect on the process of
working with a partner in this lab. What did you find
helpful about the Driver-Navigator approach? How did it
affect your understanding of the material? I think it is nice to work
with a partner, as we can build on each other’s strengths in order to
develop both of our understandings of the material. I think it was a bit
hard to do the driver-navigator approach when we have to turn in
separate documents and have to both have the code for everything. I
think it is easier to work with a partner on these activities in the
sense that we both do the code individually and then check over our work
and answers together, as you can still increase your understanding of
the material and get help where you need. —
End of Lab Activity