Presentation Ninja

# Psy202 Tutorial Week #4

July 15th 2025

### Jennet Baumbach

*We will begin @ 5:10pm*

---

## Tutorial #4 Practice Problem

> Joel was interested in whether a drug can improve running speed. Participants were recruited
into one of four conditions: no drug, placebo, low dose, or high dose. Next, they ran a 100 meter
course while being timed. Results of their time (in seconds) is below. Is there evidence to suggest
the drugs affected running speed? Make sure to conduct hypotheses, assumptions, test of
assumptions, a figure, and power if necessary. Discuss and interpret the results.

.pull-left[

``` r
data <- data.frame(
  Score = c(1,2,3,4,5),
  No_Drug = c(22,20,27,30,26),
  Placebo = c(25,23,26,34,22),
  Low_Dose = c(11,23,17,19,20),
  High_Dose = c(15,14,18,23,25)
)
```
]

.pull-right[

- I would begin by typing in the values in to a software program in order to allow for computational analysis

- Perhaps try solving the question BOTH by hand and using a software program! :)

]

---

## Tutorial #4 Practice Problem

| Score| No_Drug| Placebo| Low_Dose| High_Dose|
|-----:|-------:|-------:|--------:|---------:|
|     1|      22|      25|       11|        15|
|     2|      20|      23|       23|        14|
|     3|      27|      26|       17|        18|
|     4|      30|      34|       19|        23|
|     5|      26|      22|       20|        25|

|Stat     | No_Drug| Placebo| Low_Dose| High_Dose|
|:--------|-------:|-------:|--------:|---------:|
|mean     |      25|    26.0|       18|      19.0|
|variance |      16|    22.5|       20|      23.5|

## Psy202 Week #4 Answer

- I will solve the problem usin `Rstudio`

- Compare a "by hand" computational appraoch to build-in R formula to compute ANOVA

- First, enter the raw data:

``` r
data <- data.frame(
  Score = c(1,2,3,4,5),
  No_Drug = c(22,20,27,30,26),
  Placebo = c(25,23,26,34,22),
  Low_Dose = c(11,23,17,19,20),
  High_Dose = c(15,14,18,23,25)
)
```

---

## Psy202 Week #4 Answer

**Assumptions**

- Normality

- Independence

- Homogenaity of variances (test with F max)

`$$F_{max} = \dfrac{var_{max}}{var_{min}}$$`

``` r
Fmax = 23.5 / 16
Fmax
```

```
## [1] 1.46875
```

***Look up the critical value in the table...***

Fcrit = 20.6

.pull-left[
`$$F_{max} > F_{crit}$$`
]

.pull-right[
Therefore the homogenaity assumption is met. 
]
---

## Psy202 Week #4 Answer

.pull-left[

`$$SS_{between} = n * \Sigma (\bar{X_j} - Gm)^2$$`
]

.pull-right[

``` r
Grand_mean = mean(c(25,26,18,19))
ss_between = 5*((25-Grand_mean)^2 + (26-Grand_mean)^2 + (18-Grand_mean)^2+ (19-Grand_mean)^2)
ss_between
```

```
## [1] 250
```
]

.pull-left[

`$$df_{between} = k - 1$$`
]

.pull-right[

``` r
df_between = 4-1
df_between
```

```
## [1] 3
```

]

.pull-left[

`$$MS_{between} = \frac{SS_{between}}{df_{between}}$$`

]

.pull-right[

``` r
MS_between = ss_between / df_between
MS_between
```

```
## [1] 83.33333
```
]

---

## Psy202 Week #4 Answer

.pull-left[

`$$SS_{within} = (df_{1} * var_{1}) + (df_{2} * var_{2}) \\ + (df_{3} * var_{3})$$`

]

.pull-right[

``` r
ss_within = (4 * var(data$No_Drug)) + (4 * var(data$Placebo)) + (4 * var(data$Low_Dose) + (4 * var(data$High_Dose)))
ss_within
```

```
## [1] 328
```

]

.pull-left[

`$$df_{within} = N - k$$`
]

.pull-right[

``` r
df_within = 20-4
df_within
```

```
## [1] 16
```

]

.pull-left[
`$$MS_{within} = \frac{SS_{within}}{df_{within}}$$`
]

.pull-right[

``` r
MS_within = ss_within / df_within
MS_within
```

```
## [1] 20.5
```
]

---

## Psy202 Week #4 Answer

.pull-left[
`$$F = \frac{MS_{between}} {MS_{within}}$$`
]

.pull-right[

``` r
F = MS_between / MS_within
F
```

```
## [1] 4.065041
```

]

``` r
# Manually find critical F value for (2,18) df
F_crit = qf(p=.05,df1=df_between,df2=df_within,lower.tail=FALSE)
F_crit
```

```
## [1] 3.238872
```

- Same result as the table in the back of your textbook, I am just showing a code-based approach to get the critical value here :)

`$$F_{obtained} = 4.0650  > F_{critial} = 3.2389$$`

- Conclusion: we **reject** the null hypothesis and conclude that there **is evidence** that the four groups are not all equal.

---

## Psy202 Week #4 Answer

- Compare to build-in R formula

.pull-left[

``` r
library(reshape2)
a <- data %>% 
  melt(id.vars = "Score")

head(a)
```

```
##   Score variable value
## 1     1  No_Drug    22
## 2     2  No_Drug    20
## 3     3  No_Drug    27
## 4     4  No_Drug    30
## 5     5  No_Drug    26
## 6     1  Placebo    25
```
]

.pull-right[

``` r
a$variable <- factor(a$variable,levels = c("No_Drug","Placebo","Low_Dose","High_Dose"))

head(a)
```

```
##   Score variable value
## 1     1  No_Drug    22
## 2     2  No_Drug    20
## 3     3  No_Drug    27
## 4     4  No_Drug    30
## 5     5  No_Drug    26
## 6     1  Placebo    25
```
]
---

## Psy202 Week #4 Answer

``` r
b <- aov(data = a, value~variable)
summary(b)
```

```
##             Df Sum Sq Mean Sq F value Pr(>F)  
## variable     3    250   83.33   4.065 0.0252 *
## Residuals   16    328   20.50                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
```

- Same F statistics confirms that the by-hand calculations were correct! :)

---

## Psy202 Week #4 Answer

- Instructions say to make a figure! (F-U-N FUN!!)

.pull-left[

``` r
data %>%
  melt(id.vars = "Score") %>%
  group_by(variable) %>% 
  summarise(
    n=n(),
    mean=mean(value),
    sd = sd(value)
  ) %>% mutate(se = sd / sqrt(n)) %>%
  ggplot(aes(x=variable,y=mean,colour=variable, fill=variable))+
  geom_bar(stat="identity",alpha=0.2)+
  geom_errorbar(aes(x=variable,ymin=mean-se,ymax=mean+se),width=0.5)+
  theme_classic()+
  theme(legend.position = "none")+
  theme(plot.title = element_text(hjust = 0.5))+
  labs(
    x="Drug Conditions",
    y="Time (seconds)",
    title = "Effect of Drug X on Race Times"
  )+
  ylim(0,35)
```
]

.pull-right[
<img src="data:image/png;base64,#Graph.png" width="1600" />
]

- Data expressed as mean value +/- the standard error of the mean. 
---

## Questions? Comments? Concerns?? :)