OBJECTIVE: In our lectures this unit, we spent a good chunk of time reviewing various hypothesis tests which allow us to compare means to expected null values and determine if they fit our hypothesized expectations. For this R lab, you will be tasked with performing various types of hypothesis tests. For the first few, I will tell you WHAT kind of test you should run, for the last few, YOU will have to decide based on the type of data you’re working with and the question I am asking.

For reference, I have created a markdown file that contains example code for running all types of hypothesis tests, feel free to use those or to use internet resources.

Example 1

From the 3rd R lab…Please bring in the “rice.csv” file, you may call it whatever you want. The folks that collected the data would like you to analyze it to see what patterns may/may not exist. The first thing they want you to look at is whether or not the means of RootDryMass differs between wild-type and gmo rice. To run a comparison of means between these two groups, you will need to run a two-sample t-test. Use the code chunk below to bring in the data and to run the test.

#should you decide to use read.table()...
rice <- read.csv("rice.csv", header = TRUE)
str(rice)
## 'data.frame':    72 obs. of  7 variables:
##  $ PlantNo     : int  1 2 3 4 5 6 7 8 9 10 ...
##  $ Block       : int  1 1 1 1 1 1 2 2 2 2 ...
##  $ RootDryMass : int  56 66 40 43 55 66 41 67 40 35 ...
##  $ ShootDryMass: int  132 120 108 134 119 125 98 122 114 82 ...
##  $ trt         : chr  "F10" "F10" "F10" "F10" ...
##  $ fert        : chr  "F10" "F10" "F10" "F10" ...
##  $ variety     : chr  "wt" "wt" "wt" "wt" ...
t.test(RootDryMass ~ variety, data = rice, conf.level = 0.95, alternative = "two.sided", var.equal = T)
## 
##  Two Sample t-test
## 
## data:  RootDryMass by variety
## t = -5.2857, df = 70, p-value = 1.353e-06
## alternative hypothesis: true difference in means between group gmo and group wt is not equal to 0
## 95 percent confidence interval:
##  -23.14671 -10.46440
## sample estimates:
## mean in group gmo  mean in group wt 
##          9.666667         26.472222

  1. What are your null and alternative hypotheses for this test?

  2. What is the p-value for the two sample t test? What does this mean? #### Answer:

  3. The null hypothesis states that there is NO difference in RootDryMass means between wt and gmo rice varieties. The alternate hypothesis states that there IS a difference in RootDryMass means between wt and gmo rice varieties.

  4. The p-value is 1.353e-06, This is the probability that the results of the t.test are a result of random chance. Based on this p-value, I reject the null hypothesis.

Let’s create a boxplot comparing the RootDryMass for each variety

boxplot(RootDryMass ~ variety, 
data = rice, 
col = c("lavender", "purple"), 
main = "Root Dry Mass between Wild Type and GMO Rice Varieties")

Example 2: rice continued…

Next, the researchers want to check if other factors are influencing growth of rice. Given that they used a different fertilizer, they are curious if that impacted the RootDryMass Since there are more than 2 levels of fertilizer, the appropriate test would be an ANOVA. Use the code chunk below to run an ANOVA of how means of RootDryMass differ across fertilizer treatments (‘fert’)

# place your code here
riceANOVA <- aov(RootDryMass ~ fert,
          data = rice)
summary(riceANOVA)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## fert         2   3640  1819.8   8.855 0.000378 ***
## Residuals   69  14181   205.5                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. What are you null and alternative hypotheses for this test?
  2. What is the P-value of this test? What does this mean?
  3. Do we need to run a post-hoc test? Why or why not? a. if you DO need to run a post-hoc test, run it in the code chunk below. b. What does your post-hoc analysis tell you?

Answers:

  1. The null hypothesis states that the RootDryMass means of the 3 fertilizer types ARE the same. The alternate hypothesis states that the RootDryMass means of the 3 fertilizers are NOT the same.
  2. The p-value of this test is 0.000378. This is a significant p-value because it is less than 0.05, so we therefore reject the null hypothesis.
  3. Yes, it is important to run a post-hoc test because the ANOVA was significant, ths tells us at least one fertilizer group is significantly different, but it does not specify which one. The tukey test distinguishes which fertilizer group or groups are different. b. The post-hoc test states that mean RootDryMass is significantly different between F10 and NH4CI (p-adj = 0.00035) and F10 and NH4NO3 (p adj = 0.0123). However, the mean RootDryMass is not significantly different between NH4CI and NH4NO3 (p adj = 0.494). Therefore, the F10 fertilizer is significantly different from both of the other fertilizers, which do not differ significantly with each other.
TukeyHSD(riceANOVA)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = RootDryMass ~ fert, data = rice)
## 
## $fert
##                    diff        lwr       upr     p adj
## NH4Cl-F10    -16.875000 -26.787877 -6.962123 0.0003499
## NH4NO3-F10   -12.166667 -22.079543 -2.253790 0.0122604
## NH4NO3-NH4Cl   4.708333  -5.204543 14.621210 0.4943812

Let’s create a boxplot comparing the RootDryMass for each fert

boxplot(RootDryMass ~ fert, 
data = rice, 
col = c("blue", "skyblue", "navy"), 
main = "Root Dry Mass between Fertilizer types")

Example 2

Preventative care is incredibly important for finding and treating many disease before they become a problem. Researchers have been collecting data on one of the most common types of cancer, breast cancer, and how it associates with having preventative screening, in this case, a mammogram. Researchers have collected data on over 80,000 women over the last few decades and have been tracking whether women died from breast cancer and whether they had a mammogram. These researchers have asked you to look into possible links between the two. Here, since both variables are categorical, you will need to test for associations using a chi-squared contingency test. In the code chunk below, please read in the “mammogram.csv” dataset and run the chi-squared analysis.

Answer:

mam <- read.csv("mammogram.csv", header = TRUE)
str(mam)
## 'data.frame':    89835 obs. of  3 variables:
##  $ rownames           : int  1 2 3 4 5 6 7 8 9 10 ...
##  $ treatment          : chr  "mammogram" "mammogram" "mammogram" "mammogram" ...
##  $ breast_cancer_death: chr  "yes" "yes" "yes" "yes" ...
chisq.test(mam$treatment, mam$breast_cancer_death)
## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  mam$treatment and mam$breast_cancer_death
## X-squared = 0.01748, df = 1, p-value = 0.8948
  1. What are your null and alternative hypotheses for this test?
  2. How do you interpret the p-value for this relationship?

Answers:

  1. The null hypothesis states that there is NO association between mammogram treatment and breast cancer death. The alternate hypothesis states that there IS an association between mammogram treatment and breast cancer death.
  2. The p-value for this relationship is 0.8948, which is not a significant value, so we fail to reject the null hypothesis because there is insufficient evidence to support a claim of association. This means that there is an 89% chance that the results occured due to chance.

Let’s create a mosaic plot to show the relationship between cancer survival and whether or not a patient had a mammogram.

# place your code here
mosaicplot(~treatment + breast_cancer_death, data = mam)

****

Example 3: YOU CHOOSE THE RIGHT TEST!

NOTE: for whatever test you end up choosing, you will need to make sure you address the following: 1. tell me WHY you are choosing the test you’re choosing 2. Tell me your null and alternative hypotheses 3. Interpret the p-value and what it means for the potential relationship 4. Create an appropriate graph for the relationship. 5. If you choose an ANOVA, you need to determine whether or not you need to run a post-hoc test! Run it if you believe it is necessary.

Background: in the code chunk below, please bring in “poison.csv”, you may call the dataframe whatever you want. The data comes from a study looking at how different types of poison and various treatments (levels) of the poison impact how long an animal can survive for (days). Given the data from the experiment, the researchs want you to look at how the mean time until death differs by ‘treat’

HINT: you can run some of the initial code we looked at in the first few R labs to determine the types of data you have as well as the levels within the variables.

poison <- read.csv("poisons.csv", header = TRUE)
str(poison)
## 'data.frame':    48 obs. of  3 variables:
##  $ time  : num  0.31 0.45 0.46 0.43 0.36 0.29 0.4 0.23 0.22 0.21 ...
##  $ poison: int  1 1 1 1 2 2 2 2 3 3 ...
##  $ treat : chr  "A" "A" "A" "A" ...
table(poison$treat)
## 
##  A  B  C  D 
## 12 12 12 12
poisonANOVA <- aov(time ~ treat,
          data = poison)
summary(poisonANOVA)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## treat        3 0.9212 0.30707   6.484 0.000992 ***
## Residuals   44 2.0839 0.04736                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
TukeyHSD(poisonANOVA)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = time ~ treat, data = poison)
## 
## $treat
##            diff         lwr         upr     p adj
## B-A  0.36250000  0.12528287  0.59971713 0.0010358
## C-A  0.07833333 -0.15888380  0.31555047 0.8143113
## D-A  0.22000000 -0.01721713  0.45721713 0.0778376
## C-B -0.28416667 -0.52138380 -0.04694953 0.0131752
## D-B -0.14250000 -0.37971713  0.09471713 0.3869986
## D-C  0.14166667 -0.09555047  0.37888380 0.3921830
boxplot(
  time ~ treat,
  data = poison,
  main = "Time Until Death by Treatment",
  xlab = "Treatment",
  ylab = "Time Until Death",
  col = c("lavender", "violet", "purple", "purple4")
)

  1. I chose a One-way ANOVA because the time until death is a numerical value and treat is a categorical value with more than two groups.
  2. The null hypothesis states that mean time until death IS the same across all treatments. The alternate hypothesis states that the mean time until death IS NOT the same across all treatments.
  3. The p-value of the ANOVA is 0.000992. This is less than p-value = 0.05 which would suggest that at least one of the groups in the treatment column is significantly different from the others. Based on this analysis, I reject the null hypothesis.
  4. I chose a boxplot to plot the results.
  5. I have determined a post-hoc test is necessary because the p-value for the initial ANOVA was less than 0.05, indicating a significant difference between at least one group test without specifying which. The post-hoc analysis results suggest that groups B-A and C-B are significantly different from each other.