1

Cell and marginal means are below.

d <- read.csv("https://whlevine.hosted.uark.edu/psyc5143/avoidance.csv")

# better labels; this does some of the work for me for part f
d <- d %>% 
    mutate(areaF = ifelse(area == 1, "neutral", "A"),
                 delayF = case_when(delay == 1 ~ "50",
                                                     delay == 2 ~ "100",
                                                     delay == 3 ~ "150"))

# means
d %>% group_by(areaF, delayF) %>% summarize(M = mean(latency))

## # A tibble: 6 × 3
## # Groups:   areaF [2]
##   areaF   delayF     M
##   <chr>   <chr>  <dbl>
## 1 A       100       25
## 2 A       150       26
## 3 A       50        14
## 4 neutral 100       25
## 5 neutral 150       25
## 6 neutral 50        26

d %>% group_by(areaF) %>% summarise(M = mean(latency))

## # A tibble: 2 × 2
##   areaF       M
##   <chr>   <dbl>
## 1 A        21.7
## 2 neutral  25.3

d %>% group_by(delayF) %>% summarise(M = mean(latency))

## # A tibble: 3 × 2
##   delayF     M
##   <chr>  <dbl>
## 1 100     25  
## 2 150     25.5
## 3 50      20

See the code below. Your answers will differ somewhat if you used different contrasts than I did for the delay factor.

# contrast codes
d <- d %>% 
    mutate(areaC = ifelse(areaF == "neutral", 1/2, -1/2),
                 delayC1 = ifelse(delayF == "50", -2/3, 1/3),
                 delayC2 = case_when(delayF == "100" ~ 1/2,
                                                        delayF == "150" ~ -1/2,
                                                        TRUE ~ 0),
                 int1 = areaC*delayC1,
                 int2 = areaC*delayC2)

The intercept below is the mean of the cell means. The areaC slope is the difference between the Area A and the neutral area means. The delayC1 slope is the difference between the mean of the 100 & 150 msec conditions combined and the mean of the 50 msec condition. The delayC2 slope is the difference between the 100 and 150 msec condition means. The int1 slope is the difference 100/150 msec combined and 50 msec in Area A vs the same difference in the neutral area. The int2 slope is the difference between 100 and 150 msec in Area A vs the same difference in the neutral area.

model <- lm(latency ~ areaC + delayC1 + delayC2 + int1 + int2, d)
coef(model)

## (Intercept)       areaC     delayC1     delayC2        int1        int2 
##      23.500       3.667       5.250      -0.500     -12.500       1.000

Latencies are not significantly shorter when Area A has been lesioned, relative to the neutral area lesion, t(24) = 1.98, p = .06. Latencies are significantly longer in the 100 and 150 msec delay conditions than in the 50 msec condition, t(24) = 2.67, p = .01; this effect is significantly larger in Area A than in the neutral area, t(24) = 3.18, p = .004. There is no significant difference in latencies in the 100 and 150 msec conditions (p = .83), and this difference did not differ significantly between the two lesioned areas (p = .83).

summary(model)

anova(model)

## Analysis of Variance Table
## 
## Response: latency
##           Df Sum Sq Mean Sq F value Pr(>F)   
## areaC      1    101   100.8    3.92  0.059 . 
## delayC1    1    184   183.8    7.14  0.013 * 
## delayC2    1      1     1.2    0.05  0.827   
## int1       1    260   260.4   10.11  0.004 **
## int2       1      1     1.2    0.05  0.827   
## Residuals 24    618    25.8                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The SSRs for the predictors are above in the Sum Sq column.

See the code below.

d$areaF <- as.factor(d$areaF)
d$delayF <- as.factor(d$delayF)

summary(aov(latency ~ areaF*delayF, d))

For areaC, SSR = 100.83, which is the same as the Sum Sq value for the area factor in the ANOVA. The SSR values for delayC1 and delayC2 sum (183.75 + 1.25 = 185) to the Sum Sq value for the delay factor in the ANOVA. The SSR values for int1 and int2 sum (260.42 + 1.25 = 261.67) to the Sum Sq value for the interaction in the ANOVA.
The code below is overboard, but I want to avoid problems with rounding. What I’ve done is extracted SSR values and the effect-size measures and stashed them in a data frame. I also put SSE in a variable. The I calculated the effect size measures so that you can see them side-by-side. Yeah, it’s overboard, but it keeps me sharp.

# build a data frame by first pulling out the pieces of interest from the modelEffectSizes
d2 <- data.frame(modelEffectSizes(model)$Effects)
# and then get rid of the intercept row
d2 <- d2[-which(rownames(d2) == "(Intercept)"), ]

# I can't quit lmSupport !

# put SSE & SST in variables
SSE <- modelEffectSizes(model)$SSE
SST <- modelEffectSizes(model)$SST

# calculate effect-sizes "by hand" and add them to the data
d2 <- d2 %>% 
    mutate(partialEta = SSR / (SSR + SSE),
                 deltaR = SSR / SST)

# show off
d2

##            SSR df pEta.sqr   dR.sqr partialEta   deltaR
## areaC   100.83  1 0.140274 0.086515   0.140274 0.086515
## delayC1 183.75  1 0.229186 0.157658   0.229186 0.157658
## delayC2   1.25  1 0.002019 0.001073   0.002019 0.001073
## int1    260.42  1 0.296461 0.223438   0.296461 0.223438
## int2      1.25  1 0.002019 0.001073   0.002019 0.001073

Part j is answered here, too. See the code below.

# dummy codes
d <- d %>% 
    mutate(areaD = ifelse(areaF == "neutral", 0, 1),
                 delayD1 = ifelse(delayF == "100", 1, 0),
                 delayD2 = ifelse(delayF == "150", 1, 0),
                 intD1 = areaD*delayD1,
                 intD2 = areaD*delayD2)

modelDummy <- lm(latency ~ areaD+ delayD1 + delayD2 + intD1 + intD2, d)
coef(modelDummy)

Now the parameter estimates are about cell means. The intercept is the mean of the double-reference group (neutral-50). The areaD slope (-12) is the difference between the means of neutral-50 (26) and Area A-50 (14). The delayD1 slope about the difference between the mean of neutral-50 (26) and neutral-100 (25); the delayD2 slope is about the difference between the mean of neutral-50 (26) and neutral-150 (25). The intD1 slope is the difference between difference in the neutral-50 mean (26) and the neutral-100 mean (25) - which is -1 - and the same difference in Area A (14 - 25 = -11); and the intD2 slope is the corresponding difference in the comparison of the 50 vs 150 msec conditions.

2

ps8 <- read.csv("https://whlevine.hosted.uark.edu/psyc5143/ps8.csv", stringsAsFactors = T)

# a

# cell means & B means
ps8 %>% group_by(A, B) %>% summarise(M = mean(DV))
ps8 %>% group_by(B) %>% summarise(M = mean(DV))

# i, j, k
lm(DV ~ dummyA1*conB, ps8) %>% coef()
lm(DV ~ dummyA2*conB, ps8) %>% coef()
lm(DV ~ conA*conB, ps8) %>% coef()

Means are below.

#     A1  A2
# B1  10  14  12
# B2  12  20  16

The main effect of B is 16 - 12 = 4.
The simple effect of B at A1 is 12 - 10 = 2.
The simple effect of B at A2 is 20 - 14 = 6.
The average of the two simple effect is (6 + 2) / 2 = 4, which is the main effect.
Held constant at zero for dummyA1 means group (or level) A1.
Held constant at zero for dummyA2 means group (or level) A2.
Held constant at zero for conA means neither group A1 or A2, or, in a sense, the average of the two.
The conB slope is 2, which matches the simple effect of B at A1 from part c.
The conB slope is 6, which matches the simple effect of B at A2 from part d.
The conB slope is 4, which matches the average of the two simple effects of B, or its main effect, from part b.

MAIN EFFECTS ARE AVERAGES OF SIMPLE EFFECTS! Equals zero for a contrast-coded factor is about the average for the factor, which is zero, or the main effect!

Problem Set 8 Answer Key

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