Include your answers in this document in the sections below the rubric where I have point values indicated (1 p).

Rubric

Answer the questions with the data example.

Example: Child Health and Development Study (CHDS)

We consider data from the birth records of 680 live-born white male infants. The infants were born to mothers who reported for pre-natal care to three clinics of the Kaiser hospitals in northern California. As an initial analysis, we will examine whether maternal smoking has an effect on the birth weights of these children. To answer this question, we define 3 groups based on mother’s smoking history:

mother does not currently smoke or never smoked,

mother smoked less than one pack of cigarettes a day during pregnancy, and

mother smoked at least one pack of cigarettes a day during pregnancy.

Let \(\mu_i\) = pop mean birth weight (lb) for children in group \(i\), \((i=1,2,3)\). We wish to test \(H_0: \mu_1 = \mu_2 = \mu_3\) against \(H_A: \textrm{not } H_0\).

We read in the data, create a smoke factor variable, and plot the data by smoking group.

#### Example from the Child Health and Development Study (CHDS)# description at http://statacumen.com/teach/ADA1/ADA1_notes_05-CHDS_desc.txt# read data from websitedat_chds <-read_csv("ADA1_notes_05-CHDS.csv")

Rows: 680 Columns: 13
── Column specification ────────────────────────────────────────────────────────
Delimiter: ","
chr (5): id, m_smok, m_ppwt, p_educ, p_smok
dbl (8): c_head, c_len, c_bwt, gest, m_age, m_ht, p_age, p_ht
ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

dat_chds <- dat_chds %>%mutate(# create a factor variable based on number of cigarettes smokedsmoke =case_when(# no cigs m_smok ==0~"0 cigs"# less than 1 pack (20 cigs = 1 pack) , (m_smok >0) & (dat_chds$m_smok <20) ~"1-19 cigs"# at least 1 pack (20 cigs = 1 pack) , m_smok >=20~"20+ cigs" ) , smoke =factor(smoke) ) %>%select( c_bwt , smoke )summary(dat_chds)

c_bwt smoke
Min. : 3.300 0 cigs :381
1st Qu.: 6.800 1-19 cigs:169
Median : 7.600 20+ cigs :130
Mean : 7.516
3rd Qu.: 8.200
Max. :11.400

Plot the data in a way that compares the means. Error bars are 95% confidence intervals of the mean.

# Plot the data using ggplotlibrary(ggplot2)p <-ggplot(dat_chds, aes(x = smoke, y = c_bwt))# plot a reference line for the global mean (assuming no groups)p <- p +geom_hline(yintercept =mean(dat_chds$c_bwt),colour ="black", linetype ="dashed", size =0.3, alpha =0.5)# boxplot, size=.75 to stand out behind CIp <- p +geom_violin(width =0.5, alpha =0.25)p <- p +geom_boxplot(width =0.25, alpha =0.25)# points for observed datap <- p +geom_point(position =position_jitter(w =0.05, h =0), alpha =0.2)# diamond at mean for each groupp <- p +stat_summary(fun = mean, geom ="point", shape =18, size =4,colour ="red", alpha =0.8)# confidence limits based on normal distributionp <- p +stat_summary(fun.data ="mean_cl_normal", geom ="errorbar",width = .2, colour ="red", alpha =0.8)p <- p +labs(x ="Maternal smoking (per day)" , y ="Child birthweight (lb)" , title ="Child birthweight vs\nmaternal smoking" )print(p)

Hypothesis test

Set up the null and alternative hypotheses in words and notation.

In words: ``The population mean birthweight is different between smoking groups.’’

In notation: \(H_0: \mu_1=\mu_2=\mu_3\) versus \(H_A: \textrm{not } H_0\) (at least one pair of means differ).

Let the significance level of the test be \(\alpha=0.05\).

Compute the test statistic.

fit_c <-aov( c_bwt ~ smoke , data = dat_chds )summary(fit_c)

Df Sum Sq Mean Sq F value Pr(>F)
smoke 2 40.7 20.351 17.9 2.65e-08 ***
Residuals 677 769.5 1.137
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The \(F\)-statistic for the ANOVA is \(F = 17.9\).

Compute the \(p\)-value from the test statistic.

The p-value for testing the null hypothesis is \(p = 2.65\times 10^{-8}\).

(2 p) State the conclusion in terms of the problem.

[ANSWER HERE]

Check assumptions of the test.

Residuals are normal

Populations have equal variances.

Check whether residuals are normal.

Plot the residuals and assess whether they appear normal.

# Plot the data using ggplotdf_res <-data.frame(res = fit_c$residuals)library(ggplot2)p <-ggplot(df_res, aes(x = res))p <- p +geom_histogram(aes(y = ..density..), binwidth =0.2)p <- p +geom_density(colour ="blue")p <- p +geom_rug()p <- p +stat_function(fun = dnorm, colour ="red", args =list(mean =mean(df_res$res), sd =sd(df_res$res)))p <- p +labs(title ="ANOVA Residuals" , caption ="Blue = Kernal density curve\nRed = Normal distribution")print(p)

(1 p) Describe the plot of residuals as it relates to model assumptions.

[ANSWER HERE]

Plot the residuals versus the normal quantiles. If the residuals are normal, then the will fall on the center line and very few will be outside the error bands.

# QQ plotpar(mfrow=c(1,1))#library(car)car::qqPlot( fit_c$residuals , las =1 , id =list(n =4, cex =1) , lwd =1 , main="QQ Plot" )

[1] 9 505 170 337

(1 p) Describe the plot of residuals as it relates to model assumptions.

[ANSWER HERE]

A formal test of normality on the residuals tests the hypothesis \(H_0:\) The distribution is Normal vs \(H_1:\) The distribution is not Normal. We can test the distribution of the residuals.

Three tests for normality are reported below. I tend to like the Anderson-Darling test. Different tests have different properties, and tests that are sensitive to differences from normality in the tails of the distribution are typically more important for us (since deviations in the tails are more influential than deviations in the center).

shapiro.test(fit_c$residuals)

Shapiro-Wilk normality test
data: fit_c$residuals
W = 0.99553, p-value = 0.04758

Anderson-Darling normality test
data: fit_c$residuals
A = 0.62184, p-value = 0.1051

nortest::cvm.test(fit_c$residuals)

Cramer-von Mises normality test
data: fit_c$residuals
W = 0.091963, p-value = 0.1449

(1 p) Interpret the conclusion of the Anderson-Darling test.

[ANSWER HERE]

Check whether populations have equal variances.

Look at the numerical summaries below.

# calculate summariesdat_chds_summary <- dat_chds %>%group_by(smoke) %>%summarize(m =mean(c_bwt) , s =sd(c_bwt) , n =n() , .groups ="drop_last" ) %>%ungroup()dat_chds_summary

# A tibble: 3 × 4
smoke m s n
<fct> <dbl> <dbl> <int>
1 0 cigs 7.73 1.05 381
2 1-19 cigs 7.22 1.08 169
3 20+ cigs 7.27 1.09 130

(1 p) Interpret the standard deviations above. You may also discuss the plots of the data.

[ANSWER HERE]

Formal tests for equal variances. We can test whether the variances are equal between our three groups. This is similar to the ANOVA hypothesis, but instead of testing means we’re tesing variances. \(H_0: \sigma^2_1=\sigma^2_2=\sigma^2_3\) versus \(H_A: \textrm{not } H_0\) (at least one pair of variances differ).

## Test equal variance# assumes populations are normalbartlett.test(c_bwt ~ smoke, data = dat_chds)

Bartlett test of homogeneity of variances
data: c_bwt by smoke
Bartlett's K-squared = 0.3055, df = 2, p-value = 0.8583

# does not assume normality, requires car package#library(car)car::leveneTest(c_bwt ~ smoke, data = dat_chds)

Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 2 0.7591 0.4685
677

# nonparametric testfligner.test(c_bwt ~ smoke, data = dat_chds)

Fligner-Killeen test of homogeneity of variances
data: c_bwt by smoke
Fligner-Killeen:med chi-squared = 2.0927, df = 2, p-value = 0.3512

(1 p) Interpret the result of the appropriate test. If normality was reasonable then use Bartlett, otherwise use Levene.

[ANSWER HERE]

If the ANOVA null hypothesis was rejected, then perform follow-up Post Hoc pairwise comparison tests to determine which pairs of means are different.

There are several multiple comparison methods described in the notes. Let’s use Tukey’s Honest Significant Difference (HSD) here to test which pairs of populations differ.

EMM plot interpretation

This EMM plot (Estimated Marginal Means, aka Least-Squares Means) is only available when conditioning on one variable. The blue bars are confidence intervals for the EMMs; don’t ever use confidence intervals for EMMs to perform comparisons – they can be very misleading. The red arrows are for the comparisons among means; the degree to which the “comparison arrows” overlap reflects as much as possible the significance of the comparison of the two estimates. If an arrow from one mean overlaps an arrow from another group, the difference is not significant, based on the adjust setting (which defaults to “tukey”).

(2 p) Interpret the comparisons (which pairs differ).

[ANSWER HERE]

0 cigs - (1-19 cigs): means are (not) significant different

0 cigs - (20+ cigs): means are (not) significant different

(1-19 cigs) - (20+ cigs): means are (not) significant different

(1 p) Summarize results by ordering the means and grouping pairs that do not differ (see notes for examples).

[ANSWER HERE]

Replace this example with your results.
Example: Groups A and C differ, but B is not different from either.
(These groups are ordered by mean, so A has the lowest mean and C has the highest.)
Group A Group B Group C
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