Problem Set 3

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University of Texas, Dallas *

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6359

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Statistics

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Apr 3, 2024

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Question 1 In an observational study, the sample is drawn from a population of consenting subjects. To make meaningful inferences about the entire population, we must assume that the observed responses are independent of the factors or motivations that led individuals to consent to participate in the study. This assumption is essential for the generalizability of study findings to the broader population. Question 2 Through the inquiry of television-watching habits, doctors may potentially identify children at risk of high cholesterol, without the necessity of establishing a direct cause-and-effect relationship. While this application serves a limited purpose, no other significant applications have been identified. Question 3 R Code: # Define a list of 7 variables > variables <- c(68, 77, 82, 85, 53, 64, 71) > > # Generate combinations of 4 variables > combinations_of_4 <- combn(variables, 4) > > # Initialize a list to store pairs of groups > group_pairs <- list() > > # Iterate through the combinations > for (i in 1:ncol(combinations_of_4)) { + # Create a 4-variable group (Group A) + group_A <- combinations_of_4[, i] + + # Find the complementary 3-variable group (Group B) + group_B <- setdiff(variables, group_A) + + # Store the pair of groups in the list + group_pairs[[i]] <- list(group_A = group_A, group_B = group_B) + } > > # Print information for each group pair > for (i in 1:length(group_pairs)) { + cat("Group A:", group_pairs[[i]]$group_A, "\n") + cat("Group B:", group_pairs[[i]]$group_B, "\n\n") + + # Calculate and print the difference of means + diff_of_means <- mean(group_pairs[[i]]$group_A) - mean(group_pairs[[i]]$group_B) + cat("Difference of means:", diff_of_means, "\n\n") + + # Perform a t-test and print the p-value + t_test_result <- t.test(group_pairs[[i]]$group_A, group_pairs[[i]]$group_B) + cat("p-value:", t_test_result$p.value, "\n\n") + } Result: Group A: 68 77 82 85

Group B: 53 64 71 Difference of means: 15.33333 p-value: 0.0776307 Group A: 68 77 82 53 Group B: 85 64 71 Difference of means: -3.333333 p-value: 0.722817 Group A: 68 77 82 64 Group B: 85 53 71 Difference of means: 3.083333 p-value: 0.7821507 Group A: 68 77 82 71 Group B: 85 53 64 Difference of means: 7.166667 p-value: 0.5317921 Group A: 68 77 85 53 Group B: 82 64 71 Difference of means: -1.583333 p-value: 0.861695 Group A: 68 77 85 64 Group B: 82 53 71 Difference of means: 4.833333 p-value: 0.6493828 Group A: 68 77 85 71 Group B: 82 53 64 Difference of means: 8.916667 p-value: 0.4107864 Group A: 68 77 53 64 Group B: 82 85 71 Difference of means: -13.83333 p-value: 0.08840542 Group A: 68 77 53 71 Group B: 82 85 64 Difference of means: -9.75

p-value: 0.3036934 Group A: 68 77 64 71 Group B: 82 85 53 Difference of means: -3.333333 p-value: 0.7788381 Group A: 68 82 85 53 Group B: 77 64 71 Difference of means: 1.333333 p-value: 0.8787647 Group A: 68 82 85 64 Group B: 77 53 71 Difference of means: 7.75 p-value: 0.4325919 Group A: 68 82 85 71 Group B: 77 53 64 Difference of means: 11.83333 p-value: 0.2289157 Group A: 68 82 53 64 Group B: 77 85 71 Difference of means: -10.91667 p-value: 0.1934452 Group A: 68 82 53 71 Group B: 77 85 64 Difference of means: -6.833333 p-value: 0.4624748 Group A: 68 82 64 71 Group B: 77 85 53 Difference of means: -0.4166667 p-value: 0.9707535 Group A: 68 85 53 64 Group B: 77 82 71 Difference of means: -9.166667 p-value: 0.277984

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Group A: 68 85 53 71 Group B: 77 82 64 Difference of means: -5.083333 p-value: 0.5748192 Group A: 68 85 64 71 Group B: 77 82 53 Difference of means: 1.333333 p-value: 0.9027263 Group A: 68 53 64 71 Group B: 77 82 85 Difference of means: -17.33333 p-value: 0.01480113 Group A: 77 82 85 53 Group B: 68 64 71 Difference of means: 6.583333 p-value: 0.4396535 Group A: 77 82 85 64 Group B: 68 53 71 Difference of means: 13 p-value: 0.1415435 Group A: 77 82 85 71 Group B: 68 53 64 Difference of means: 17.08333 p-value: 0.0377915 Group A: 77 82 53 64 Group B: 68 85 71 Difference of means: -5.666667 p-value: 0.5290758 Group A: 77 82 53 71 Group B: 68 85 64 Difference of means: -1.583333 p-value: 0.867988 Group A: 77 82 64 71 Group B: 68 85 53

Difference of means: 4.833333 p-value: 0.6659559 Group A: 77 85 53 64 Group B: 68 82 71 Difference of means: -3.916667 p-value: 0.6562758 Group A: 77 85 53 71 Group B: 68 82 64 Difference of means: 0.1666667 p-value: 0.9854893 Group A: 77 85 64 71 Group B: 68 82 53 Difference of means: 6.583333 p-value: 0.5357546 Group A: 77 53 64 71 Group B: 68 82 85 Difference of means: -12.08333 p-value: 0.1638911 Group A: 82 85 53 64 Group B: 68 77 71 Difference of means: -1 p-value: 0.9073876 Group A: 82 85 53 71 Group B: 68 77 64 Difference of means: 3.083333 p-value: 0.7241928 Group A: 82 85 64 71 Group B: 68 77 53 Difference of means: 9.5 p-value: 0.3306254 Group A: 82 53 64 71 Group B: 68 77 85 Difference of means: -9.166667 p-value: 0.2940754

Group A: 85 53 64 71 Group B: 68 77 82 Difference of means: -7.416667 p-value: 0.3911142 Question 4 Part a Mean Calculation:  Sum of data: -6 + 0 + 1 + 2 - 3 - 4 + 2 = -8  Number of data points: 7  Mean = (-8) / 7 ≈ -1.143 Variance Calculation:  Sum of squared differences = [(-6 - (-1.143))^2 + (0 - (-1.143))^2 + (1 - (-1.143))^2 + (2 - (- 1.143))^2 + (-3 - (-1.143))^2 + (-4 - (-1.143))^2 + (2 - (-1.143))^2] ≈ 68.857  Variance = Sum of squared differences / (n - 1) = 68.857 / 6 ≈ 11.476 Standard Deviation Calculation:  Standard Deviation = √(Variance) = √11.476 ≈ 3.38 Degrees of Freedom:  Number of data points minus one: 7 - 1 = 6 Part b Standard Error = Standard Deviation/(n)^(1/2) = 1.18 Part c  Degrees of freedom (df) = 6  t-value for a 95% confidence interval with df = 6, denoted as t(α/2, df), is approximately 2.447 For a dataset with a mean (μ) of -1.143 and a standard error (σ/√n) of 1.18, we can calculate the confidence interval as follows: Lower Limit (LL) = μ + t(α/2, df) * (σ/√n) Upper Limit (UL) = μ - t(α/2, df) * (σ/√n) Substituting the values: LL = -1.143 + 2.447 * 1.18 = 1.74 UL = -1.143 - 2.447 * 1.18 = -4.03

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Part d t = (μ - μ₀) / (σ/√n) Where:  μ₀ is the hypothesized mean (in this case, 0)  μ is the sample mean (-1.143)  σ is the standard error (1.18)  n is the sample size Substituting these values: t = (-1.143 - 0) / 1.18 = -0.96 Code for p-value calculation: # Define the numeric vector > numSet <- c(-6, 0, 1, 2, -3, -4, 2) > > # Perform a one-sample t-test with a null hypothesis of mean(mu) equal to 0 > resSet <- t.test(numSet, mu = 0) > > # Extract and print the p-value from the t-test result > p_value <- resSet$p.value > cat("p-value:", p_value, "\n") p-value: 0.3790617 Question 5 Part a The null and alternative hypotheses are: Assume the level of the significance, a = 0.05 If the P-value is less than the level of the significance, then the null hypothesis should be rejected. Part b # Sample data > Year <- c(1938, 1939, 1940, 1941, 1942, 1943, 1944, 1945, 1946, 1947, 1948, 1949, 1950, 1951, 1952, 1953, 1954, 1955, 1956, 1957, 1958, 1959, 1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972) > Rate <- c(0.8, 1.3, 1.4, 1.2, 1.7, 1.8, 1.6, 1.5, 1.5, 2.0, 2.5, 2.7, 2.9, 2.5, 3.1, 2.4, 2.2, 2.9, 2.5, 2.6, 3.2, 3.8, 4.2, 3.9, 3.7, 3.3, 3.7, 3.9, 4.1, 3.8, 4.7, 4.4, 4.8, 4.8, 4.8) > Code <- c(2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1) > skin_cancer_data <- data.frame(Year,Rate,Code) > # Create two separate data frames for each group > group1_data <- data.frame(Year[Code == 1], Rate[Code == 1]) > group2_data <- data.frame(Year[Code == 2], Rate[Code == 2]) >

> # Perform a two-sample t-test > t_test_result <- t.test(group1_data$Rate, group2_data$Rate) > > # Extract the t-statistic and two-sided p-value > t_statistic <- t_test_result$statistic > t_statistic t 1.086743 > two_sided_p_value <- t_test_result$p.value > two_sided_p_value [1] 0.2863514 Since the calculated p-value exceeds the predetermined significance level, we do not have sufficient evidence to reject the null hypothesis. As a result, we can infer that employing two independent sample t-tests to compare skin cancer rates in the two groups is an appropriate approach, and there is no substantial indication of inappropriateness. Part c library(ggplot2) > # Filter the data for group 1 (Code == 1) > group1_data <- subset(skin_cancer_data, Code == 1) > # Create a scatterplot for group 1 > scatterplot_group1 <- ggplot(data = group1_data, aes(x = Year, y = Rate)) + + geom_point() + + labs(title = "Skin Cancer Rates vs. Year (Group 1)", + x = "Year", + y = "Skin Cancer Rate") > > # Display the scatterplot > print(scatterplot_group1) Group 2: library(ggplot2) > # Filter the data for group 2 (Code == 2)

> group2_data <- subset(skin_cancer_data, Code == 2) > # Create a scatterplot for group 2 > scatterplot_group2 <- ggplot(data = group2_data, aes(x = Year, y = Rate)) + + geom_point() + + labs(title = "Skin Cancer Rates vs. Year (Group 2)", + x = "Year", + y = "Skin Cancer Rate") > > # Display the scatterplot > print(scatterplot_group2) Question 6 Part a # Given data > unseeded <- c(1202.6, 830.1, 372.4, 345.5, 321.2, 244.3, 163.0, 147.8, 95.0, 87.0, 81.2, 68.5, 47.3, 41.1, 36.6, 29.0, 28.6, 26.3, 26.0, 24.4, 21.4, 17.3, 11.5, 4.9, 4.9, 1.0) > > # Create variables var_100, var_200, var_300, and var_400 > var_100 <- unseeded + 100 > var_200 <- unseeded + 200 > var_300 <- unseeded + 300 > var_400 <- unseeded + 400 > > # Combine all variables for the boxplot > all_data <- data.frame(unseeded, var_100, var_200, var_300, var_400) > > # Create a boxplot for all variables > boxplot(all_data, col = c("red", "blue", "green", "purple", "orange"), + main = "Boxplot for unseeded, var_100, var_200, var_300, var_400", + xlab = "Variables", ylab = "Values")

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Part b # Create variables var_2, var_3, var_4, and var_5 > var_2 <- unseeded * 2 > var_3 <- unseeded * 3 > var_4 <- unseeded * 4 > var_5 <- unseeded * 5 > > # Combine all variables for the boxplot > all_data <- data.frame(unseeded, var_2, var_3, var_4, var_5) > > # Create a boxplot for all variables > boxplot(all_data, col = c("red", "blue", "green", "purple", "orange"), + main = "Boxplot for unseeded, var_2, var_3, var_4, var_5", + xlab = "Variables", ylab = "Values")

Part c The additive property will be close to the actual data.

Problem Set 3

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