#R Markdown Assignment 6
#Question1a
# data
person <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
aspirin_A <- c(15, 26, 13, 28, 17, 20, 7, 36, 12, 18)
aspirin_B <- c(13, 20, 10, 21, 17, 22, 5, 30, 7, 11)
#data frame
aspirin_data <- data.frame(person, aspirin_A, aspirin_B)
aspirin_data## person aspirin_A aspirin_B
## 1 1 15 13
## 2 2 26 20
## 3 3 13 10
## 4 4 28 21
## 5 5 17 17
## 6 6 20 22
## 7 7 7 5
## 8 8 36 30
## 9 9 12 7
## 10 10 18 11Null Hypothesis (H₀): The mean concentrations of the two drugs are the same in urine specimens
Alternative Hypothesis (H₁): The mean concentrations of the two drugs are not the same in urine specimens
\(H_0: \mu_A = \mu_B\) and \(H_1: \mu_A \neq \mu_B\)
#Question1b
#performing paired t-test
paired_test <- t.test(aspirin_A, aspirin_B, paired = TRUE)
paired_test##
## Paired t-test
##
## data: aspirin_A and aspirin_B
## t = 3.6742, df = 9, p-value = 0.005121
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
## 1.383548 5.816452
## sample estimates:
## mean difference
## 3.6#extracting the p-value
p_value_paired <- paired_test$p.value
p_value_paired## [1] 0.005121073The p-value from the paired t-test is 0.0051
With α = 0.05, since the p-value is 0.0051, we reject the null hypothesis
#Conclusion: There is a significant difference between the mean concentrations of the two aspirin types
#Question1c
#performing the two-sample t-test for comparison
two_sample_test <- t.test(aspirin_A, aspirin_B, paired = FALSE)
two_sample_test##
## Welch Two Sample t-test
##
## data: aspirin_A and aspirin_B
## t = 0.9802, df = 17.811, p-value = 0.3401
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -4.12199 11.32199
## sample estimates:
## mean of x mean of y
## 19.2 15.6#extracting the p-value
p_value_two_sample <- two_sample_test$p.value
p_value_two_sample## [1] 0.3401157#comparing the p-value: The p-value from the two-sample t-test would have been 0.3401
#Comparison: The two-sample t-test gives a larger p-value (0.3401) compared to the paired t-test (0.0051)
#Question2a
#data
active_exercise <- c(9.50, 10.00, 9.75, 9.75, 9.00, 13.0)
no_exercise <- c(11.50, 12.00, 13.25, 11.50, 13.00, 9.00)
#data frame
walking_data <- data.frame(active_exercise = c(active_exercise, rep(NA, length(no_exercise)-length(active_exercise))), no_exercise = no_exercise)
walking_data## active_exercise no_exercise
## 1 9.50 11.50
## 2 10.00 12.00
## 3 9.75 13.25
## 4 9.75 11.50
## 5 9.00 13.00
## 6 13.00 9.00Null Hypothesis (H₀): The mean time to walk is the same for children who receive active exercise versus no exercise
Alternative Hypothesis (H₁): The mean time to walk is less for children who receive active exercise versus no exercise
\(H_0: \mu_{Active} = \mu_{No}\) and \(H_1: \mu_{Active} < \mu_{No}\)
#Question 2b We might want to use a non-parametric method because we have a small sample size (only 6 infants per group) which may not meet the normality assumptions required for t-tests. Non-parametric tests like the mann-whitney u test are more robust to outliers and non-normal distributions since they don’t assume the data follows a normal distribution
#Question 2c
#performing the mann-whitney u test using the wilcox test function
mann_whitney_test <- wilcox.test(active_exercise, no_exercise,alternative = "less",exact = FALSE)
mann_whitney_test##
## Wilcoxon rank sum test with continuity correction
##
## data: active_exercise and no_exercise
## W = 9, p-value = 0.08523
## alternative hypothesis: true location shift is less than 0#extracting the p-value
p_value_mann_whitney <- mann_whitney_test$p.value
p_value_mann_whitney## [1] 0.08523293#comparing the p-value: The p-value from the mann-whitney u test is 0.0852
With α = 0.05, since the p-value is 0.0852, we fail to reject the null hypothesis
Conclusion: There is not a sufficient amount evidence to conclude that active exercise reduces the mean time to walk