You are a botanist looking at variation in flower morphology along the central coast and of special interest to you is a species of native sunflower. You think that there's a difference in the number of petals between some populations: population1 and population 2. You believe that population 2 has more petals. You count the number of petals in both populations and get the data.
a.) write a properly worded hypothesis and prediction for this study.
b.) Do the populations have different numbers of petals? Use Rstudio to perform an appropriate statistical test and write a properly formatted results statement.
Data:
| Population | Petals |
| Pop 1 | 1 |
| Pop 1 | 29 |
| Pop 1 | 23 |
| Pop 1 | 30 |
| Pop 1 | 26 |
| Pop 1 | 26 |
| Pop 1 | 5 |
| Pop 1 | 14 |
| Pop 1 | 11 |
| Pop 1 | 3 |
| Pop 1 | 8 |
| Pop 1 | 15 |
| Pop 1 | 14 |
| Pop 1 | 22 |
| Pop 1 | 28 |
| Pop 2 | 39 |
| Pop 2 | 11 |
| Pop 2 | 10 |
| Pop 2 | 30 |
| Pop 2 | 26 |
| Pop 2 | 31 |
| Pop 2 | 19 |
| Pop 2 | 15 |
| Pop 2 | 22 |
| Pop 2 | 33 |
| Pop 2 | 32 |
| Pop 2 | 16 |
| Pop 2 | 32 |
| Pop 2 | 33 |
| Pop 2 | 19 |
a)
Null and Alternate Hypothesis
H0: µ2 = µ1(Mean Petals count of Population 2 is same as that of Population 1)
Ha: µ2 > µ1 (Mean Petals count of Population 2 is greater than that of Population 1)
b)
Code:
# Read Data
setwd ("D:/Data")
mydata <- read.csv("Data_23Nov19.csv", header = T)
head(mydata)
pop1 = mydata[mydata$Population=="Pop 1",]
pop2 = mydata[mydata$Population=="Pop 2",]
t.test(pop2$Petals,pop1$Petals, paired = FALSE,alternative = "greater")
Output
Welch Two Sample t-test
data: pop2$Petals and pop1$Petals
t = 2.1594, df = 27.778, p-value = 0.0198
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
1.59719 Inf
sample estimates:
mean of x mean of y
24.53333 17.00000
Result
Since the p-value is less than 0.05, we reject the null hypothesis in favour of alternate hypothesis.
Conclusion
Mean Petals count of Population 2 is greater than that of Population 1
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