Collecting Data for a group of students in a statistics

Full Answer Section

Plug in the values:

p(x) = e^(-7 + (0.06 * 50) + (1 * 3.5)) / (1 + e^(-7 + (0.06 * 50) + (1 * 3.5))) ≈ 0.997

The student has a very high estimated probability (almost certain) of getting an A with these values.

(2) Hours Needed for 50% Chance of A with GPA 3.4

We can use the log-odds formula and solve for X₁ (hours studied):

log(p(x)) = β₀ + β₁X₁ + β₂X₂
log(0.5) = -7 + 0.06X₁ + (1 * 3.4)  // p(x) is 50% chance of A
X₁ ≈ 81.6 hours

The student with a GPA of 3.4 would need to study approximately 81.6 hours to have a 50% chance of getting an A.

**2. Weekly Data Analysis (Note: R code snippets are included for reference, but results may vary slightly depending on software version or environment.)

(3) Summary and Scatterplots

library(ISLR)
summary(Weekly)
pairs(Weekly)

• Summary: This provides basic statistics like mean, median, minimum, and maximum for each variable.
• Scatterplots: These visualize pairwise relationships between variables.

Look for correlations between Year and Volume in the summary output and scatterplot matrix.

(4) Logistic Regression with Lags and Volume

model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Volume, data = Weekly)
summary(model)

Examine the p-values in the summary output. Statistically significant predictors will have low p-values (typically < 0.05).

(5) Confusion Matrix and Evaluation

predictions <- predict(model, type = "response")
cut_off <- 0.5  # Threshold for classifying up/down
cm <- table(Weekly$Direction, predictions > cut_off)
accuracy <- sum(diag(cm)) / sum(cm)

precision <- cm[1, 1] / (cm[1, 1] + cm[1, 2])
recall <- cm[1, 1] / (cm[1, 1] + cm[2, 1])

print(cm)
cat("Accuracy:", accuracy, "
")
cat("Precision:", precision, "
")
cat("Recall:", recall, "
")

• Confusion Matrix: This shows how well the model classified up/down movements (actual vs. predicted).
• Accuracy: Proportion of correctly predicted observations.
• Precision: Proportion of true positives among predicted positives.
• Recall: Proportion of true positives identified by the model.

(6) Logistic Regression with Lag 2 for Held-Out Data (2010)

train_data <- Weekly[Weekly$Year < 2010, ]
test_data <- Weekly[Weekly$Year == 2010, ]

model_lag2 <- glm(Direction ~ Lag2, data = train_data)
predictions_lag2 <- predict(model_lag2, newdata = test_data, type = "response")

cm_lag2 <- table(test_data$Direction, predictions_lag2 > cut_off)
accuracy_lag2 <- sum(diag(cm_lag2)) / sum(cm_lag2)

precision_lag2 <- cm_lag2[1, 1] / (cm_lag2[1, 1] + cm_

Sample Solution

1. Logistic Regression Calculations

(1) Probability of an A with 50 Hours Studied and 3.5 GPA

We can estimate the probability (p(x)) using the logistic regression formula:

p(x) = e^(β₀ + β₁X₁ + β₂X₂) / (1 + e^(β₀ + β₁X₁ + β₂X₂))

where:

• β₀ = -7 (estimated coefficient)
• β₁ = 0.06 (estimated coefficient)
• β₂ = 1 (estimated coefficient)
• X₁ = 50 (hours studied)
• X₂ = 3.5 (undergrad GPA)