(10 points)
Recall the Peachtree road race data that we have for 2010. You can read it into an R dataframe like so:
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
Select a random age n between 25 and 55 and run an ANOVA test to see if runners speeds change over the ages from n to n+5. Thus, in our last example on our ANOVA outline page, the ages variable might be generated by
set.seed(0)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
Here is my code for the ANOVA Test.
Code
df=read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
ages = 25:55
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
mod1 = lm(Net.Time ~ Age, data = df2)
anova(mod1)
Output
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 30 204276 6809.2 15.483 < 2.2e-16 ***
Residuals 37836 16639296 439.8
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Conclusion
H_o: Runners speed does not change over age.
H_A: Runners speed does change over age.
With this given p-value the null hypothesis is rejected. The the speed of the runners do change over age.
Here is how I did things:
Code
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(26)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
modl = lm(Net.Time ~ Age, data = df2)
anova(modl)
That code gave me this summary table:
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 6474 1294.85 3.1188 0.008154 **
Residuals 6558 2722702 415.17
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Conclusion:
Ho: μ1=μ2
Ha: at least two μ’s are different
calculated p-value: 0.008154
At a 95% level of confidence, we are able to reject the null.
Code:
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(3)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
ages = 25:35
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
modl = lm(Net.Time ~ Age, data = df2)
anova(modl)
Response:
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 10 8808 880.83 2.0892 0.02194
Residuals 12821 5405583 421.62
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1*
Conclusion:
Since the calculated p-value is 0.02194, we are at a 95% confidence level to reject the null hypothesis.
Input
df=read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(21)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
runners=subset(df,Age %in%ages)
runners$Age=factor(runners$Age, labels = ages)
runners.mod1= lm(Net.Time~Age,data = runners)
anova(runners.mod1)
Output
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 3350 669.99 1.4283 0.2105
Residuals 7134 3346496 469.09
Conclusion
H_0: μ_1=μ_2 Running speed stays the same over the age span
H_a: u_1 ≠ u_2 Running speeds will not stay the same over the age span.
P-value of 0.2105 is > 0.05. I fail to reject the null hypothesis at the 95% confidence level.
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(1)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
ages = 25:35
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
modl = lm(Net.Time ~ Age, data = df2)
anova(modl)
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 10 8808 880.83 2.0892 0.02194 *
Residuals 12821 5405583 421.62
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
With a p-value of .02194 and a 95% confidence level, I reject the null.
Code Used
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(14)
start_age = sample(27:48, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
mod1 = lm(Net.Time ~ Age, data = df2)
anova(mod1)
Output
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 843 168.51 0.3954 0.8523
Residuals 7581 3230426 426.12
Interpretation
There is no difference in speed from n to n+5, p-value 0.8523. Therefore professor will not be any slower next year.
Code
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(9)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
mod1 = lm(Net.Time ~ Age, data = df2)
anova(mod1)
OutPut
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 802 160.30 0.3759 0.8656
Residuals 7470 3185914 426.49
Interpretation
At a 95% confidence level with a p-value of 0.8656 we would fail to reject the null hypothesis.
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(32)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 <- subset(df, Age %in% ages)
df2$Age <- factor(df2$Age, labels = ages)
df2.mod1 <- lm(Net.Time ~ Age, data = df2)
anova(df2.mod1)
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 4097 819.48 1.8605 0.09773 .
Residuals 8088 3562558 440.47
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
with a p-value of 0.09773 we can reject the null hypothesis and confirm that runners’ times do change from ages 32 to 37
code:
set.seed(0)
start_age = 36
ages = start_age:(start_age+5)
set2 = subset(df, age %in% ages)
set2 = subset(df, Age %in% ages)
set2$Age = factor(set2$Age, labels = ages)
mod = lm(Net.Time ~ Age,data= set2)
anova(mod)
Return:
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 5106 1021.29 2.3648 0.03738 *
Residuals 7552 3261540 431.88
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
H_O: μ_1=μ_2
H_A:μ_1 ≠ μ_2
At a 95% confidence level, we reject the null hypothesis due to a p-value of 0.03738
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(12)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels =ages)modl = lm(Net.Time ~ Age, data = df2)anova(modl)
Output:
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 3339 667.70 1.609 0.1539
Residuals 7195 2985803 414.98
My P-value is 0.1539 is small enough to accept the null.
Code
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(13)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
modl = lm(Net.Time ~ Age, data = df2)
anova(modl)
Output
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 2087 417.32 0.8947 0.4835
Residuals 7415 3458485 466.42
Statement of Hypothesis
H_0: μ_1=μ_2
H_a: At least 2 of μ's are different
The output of the code generated a P-value of 0.4835.
0.4835 > 0.05
So at a 95% confidence level we fail to reject the null hypothesis.
**Code**
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(11)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
mod1 = lm(Net.Time ~ Age, data = df2)
anova(mod1)
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 650 129.95 0.3005 0.9128
Residuals 7618 3294407 432.45
**Hypothesis**
That over time there speed did not change.
Conclusion
With the given P-value the hypothesis is rejected.
Code
df = read.csv('https://www.marksmath.org/data/peach_tree2015.csv')
set.seed(28)
start_age = sample(25:55, 1)
ages = start_age:(start_age+5)
df2 = subset(df, Age %in% ages)
df2$Age = factor(df2$Age, labels = ages)
modl = lm(Net.Time ~ Age, data = df2)
anova(modl)
Output
Analysis of Variance Table
Response: Net.Time
Df Sum Sq Mean Sq F value Pr(>F)
Age 5 6474 1294.85 3.1188 0.008154 **
Residuals 6558 2722702 415.17
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Conclusion
H_0 = μ_1=μ_2
H_a = At least two μ's are different.
p-value is less than .05, therefore we reject the null.
With a 95% confidence level??
Keep in mind - the smaller the p-value, the more likely you are to reject the null. 
Which hypothesis? There’s more than one.