diff_months: 9

Australian Population Assignment

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Added on: 2023-08-18 10:34:45
Order Code: CLT304395
Question Task Id: 0
  • Country :

    Australia

Q1.

Given percentages and sample size

percentage_hot_chocolate <- 6 / 100 sample_size <- 309

Calculate the expected number of students who prefer hot chocolate

expected_hot_chocolate <- percentage_hot_chocolate * sample_size

Print the result

print(expected_hot_chocolate)q2

Given student sample data

student_sample <- c(45, 18, 61)

Given Australian population proportions

australian_population <- c(0.46, 0.25, 0.29)

Calculate expected frequencies under the assumption of independence

expected_frequencies <- sum(student_sample) * australian_population

Calculate the chi-squared statistic

chi_squared <- sum((student_sample - expected_frequencies)^2 / expected_frequencies)

Print the chi-squared distance

print(chi_squared)

03a

Given sample data

sample_size <- 261 preferences_per_plan <- c(110, 85, 66)

Calculate the expected number of preferences per plan under uniform distribution

expected_per_plan <- sample_size / length(preferences_per_plan)

Print the expected number of preferences per plan

print(expected_per_plan)

q3b

Given observed preferences and sample size

observed_preferences <- c(110, 85, 66) sample_size <- 261

Calculate the expected number of preferences per plan under uniform distribution

expected_per_plan <- sample_size / length(observed_preferences)

Calculate the chi-squared statistic

chi_squared <- sum((observed_preferences - expected_per_plan)^2 / expected_per_plan)

Calculate the degrees of freedom (number of categories - 1)

df <- length(observed_preferences) - 1

Calculate the p-value using chi-squared distribution

p_value <- 1 - pchisq(chi_squared, df)

Print the test statistic and p-value

cat(“Chi-squared statistic:”, chi_squared, “”) cat(“Degrees of freedom:”, df, “”) cat(“P-value:”, p_value, “”)

q4

Given observed data

observed_data <- matrix(c(55, 37, 37, 71, 53, 28), ncol = 3, byrow = TRUE) rownames(observed_data) <- c(“HR”, “Marketing”) colnames(observed_data) <- c(“Public Transport”, “Driving”, “Bike”)

Calculate row and column totals

row_totals <- rowSums(observed_data) col_totals <- colSums(observed_data)

Calculate the total number of employees

total_employees <- sum(row_totals)

Calculate the proportion of employees who drive

proportion_drive <- col_totals[“Driving”] / total_employees

Calculate the expected number of employees in Marketing who drive

expected_drive_marketing <- proportion_drive * row_totals[“Marketing”]

Print the expected number of employees in Marketing who drive

cat(“Expected number of employees in Marketing who drive:”, expected_drive_marketing, “”)

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  • Uploaded By : Mohit
  • Posted on : August 18th, 2023
  • Downloads : 0
  • Views : 134

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