This homework has two parts. Part 1 asks you to write your own functions. Part 2 applies hypothesis tests to two real scenarios.


Part 1 — Functions

# Q1. Write a function called calculate_area_of_rectangle that takes two parameters
#     (length, width) and returns the area (area = length * width).
#     Test it with 2 different inputs.

# Function to calculate area
calculate_area_of_rectangle <- function(length, width) {
  area <- length * width
  return(area)
}

# Test the function
calculate_area_of_rectangle(5, 3)
## [1] 15
calculate_area_of_rectangle(10, 4)
## [1] 40
# Q2. Write a function called calculate_average that takes a numeric vector and returns
#     its average. Handle the case of an empty vector by printing a message.
#     (Hint: use if/else and length(x) == 0)

calculate_average <- function(x) {

  if (length(x) == 0) {
    print("Vector is empty.")
  } else {
    return(mean(x))
  }

}

# Test
calculate_average(c(4, 8, 12, 16))
## [1] 10
calculate_average(c())
## [1] "Vector is empty."
# Q3. Write a function called check_even_odd that takes an integer and prints whether
#     it is "Even" or "Odd".
#     Test it on 14 and 27.
#     (Hint: use the %% modulus operator)

check_even_odd <- function(n) {

  if (n %% 2 == 0) {
    print("Even")
  } else {
    print("Odd")
  }

}

# Test
check_even_odd(14)
## [1] "Even"
check_even_odd(27)
## [1] "Odd"

Part 2 — Hypothesis Testing

Problem 1 — Two-Proportion z-test

In 2017, of the 144,790 students who took the AP Biology exam, 84,200 were female. That same year, of the 211,693 students who took the AP Calculus AB exam, 102,598 were female.

Is there enough evidence to show that the proportion of female students taking the Biology exam is HIGHER than the proportion taking the Calculus AB exam? Test at the 5% level.

State your hypotheses:

  • H₀: p₁ ____ p₂
  • H₁: p₁ ____ p₂
# Q4. Run the appropriate two-proportion test.
#     (Hint: prop.test(c(84200, 102598), c(144790, 211693), alternative = "greater"))

prop.test(
  c(84200, 102598),
  c(144790, 211693),
  alternative = "greater",
  correct = FALSE
)
## 
##  2-sample test for equality of proportions without continuity correction
## 
## data:  c(84200, 102598) out of c(144790, 211693)
## X-squared = 3235.3, df = 1, p-value < 2.2e-16
## alternative hypothesis: greater
## 95 percent confidence interval:
##  0.09409523 1.00000000
## sample estimates:
##    prop 1    prop 2 
## 0.5815319 0.4846547
# Q5. What is the p-value? At α = 0.05, do you reject H₀?
#Since

#p-value<0.05,

#we reject H₀.

Q6. Write your conclusion in plain English (one or two sentences): There is strong evidence that the proportion of female students taking the AP Biology exam is higher than the proportion of female students taking the AP Calculus AB exam. —

Problem 2 — Paired t-test

A vitamin K shot is given to infants soon after birth. Researchers want to see if how the infants are handled can reduce the pain. They measured how long (in seconds) the infant cried after the shot. One group received the shot the conventional way; the other group received it while the mother held the infant.

Is there enough evidence to show that infants cried LESS on average when held by their mothers vs. the conventional method? Test at the 5% level.

Old <- c(63, 0, 2, 46, 33, 33, 29, 23, 11, 12, 48, 15, 33, 14, 51,
         37, 24, 70, 63, 0, 73, 39, 54, 52, 39, 34, 30, 55, 58, 18)

New <- c(0, 32, 20, 23, 14, 19, 60, 59, 64, 64, 72, 50, 44, 14, 10,
         58, 19, 41, 17, 5, 36, 73, 19, 46, 9, 43, 73, 27, 25, 18)

State your hypotheses:

  • H₀: μ_d ____ 0
  • H₁: μ_d ____ 0
# Q7. Run a paired t-test.
#     (Hint: t.test(Old, New, paired = TRUE))

t.test(Old, New,
       paired = TRUE,
       alternative = "greater")
## 
##  Paired t-test
## 
## data:  Old and New
## t = 0.028519, df = 29, p-value = 0.4887
## alternative hypothesis: true mean difference is greater than 0
## 95 percent confidence interval:
##  -9.762971       Inf
## sample estimates:
## mean difference 
##       0.1666667
# Q8. What is the p-value? At α = 0.05, do you reject H₀?
#The p-value is approximately 0.489.

#Since
# 0.489>0.05,we fail to reject H₀.

Q9. Write your conclusion in plain English. Does the data support the claim that the new method reduces crying time? The data do not provide enough evidence to conclude that infants cried less when they were held by their mothers during the vitamin K shot. Based on this sample, the new method does not appear to significantly reduce crying time.