Homework 2

Code

library(readxl)
library(dplyr)
library(magrittr)
knitr::opts_chunk$set(echo = TRUE)

Questions:

1.

Code

# Set sample size, mean, and standard deviation
n_bypass <- 539
mean_wait_time_bypass <- 19
sd_bypass <- 10

# Calculate the standard error
se_bypass <- sd_bypass / sqrt(n_bypass)

# Calculate the margin of error for a 90% confidence interval
me_bypass <- qnorm(0.95) * se_bypass

# Calculate the lower and upper bounds of the confidence interval
lower_bypass <- mean_wait_time_bypass - me_bypass
upper_bypass <- mean_wait_time_bypass + me_bypass

# Print the results
cat("90% confidence interval for Bypass: [", round(lower_bypass, 3), ",", round(upper_bypass, 3), "]\n")

90% confidence interval for Bypass: [ 18.292 , 19.708 ]

Code

# Set sample size, mean, and standard deviation
n_angiography <- 847
mean_wait_time_angiography <- 18
sd_angiography <- 9

# Calculate the standard error
se_angiography <- sd_angiography / sqrt(n_angiography)

# Calculate the margin of error for a 90% confidence interval
me_angiography <- qnorm(0.95) * se_angiography

# Calculate the lower and upper bounds of the confidence interval
lower_angiography <- mean_wait_time_angiography - me_angiography
upper_angiography <- mean_wait_time_angiography + me_angiography

# Print the results
cat("90% confidence interval for Angiography: [", round(lower_angiography, 3), ",", round(upper_angiography, 3), "]\n")

90% confidence interval for Angiography: [ 17.491 , 18.509 ]

Code

# Create a sequence of x values
x <- seq(min(lower_bypass, lower_angiography) - 1, max(upper_bypass, upper_angiography) + 1, length = 100)

# Create normal probability density functions for each dataset
y_bypass <- dnorm(x, mean_wait_time_bypass, sd_bypass)
y_angiography <- dnorm(x, mean_wait_time_angiography, sd_angiography)

# Plot the normal density curves
plot(x, y_bypass, type = "l", col = "blue", ylim = c(0, max(y_bypass, y_angiography)), main = "Confidence Intervals Comparison", xlab = "x", ylab = "Density")
lines(x, y_angiography, type = "l", col = "red")

# Draw vertical lines at the lower and upper bounds of the confidence intervals for each dataset
abline(v = lower_bypass, col = "blue")
abline(v = upper_bypass, col = "blue")
abline(v = lower_angiography, col = "red")
abline(v = upper_angiography, col = "red")

# Draw shaded areas for the confidence intervals for each dataset
polygon(c(lower_bypass, x[x >= lower_bypass & x <= upper_bypass], upper_bypass), c(0, y_bypass[x >= lower_bypass & x <= upper_bypass], 0), col = "lightblue")
polygon(c(lower_angiography, x[x >= lower_angiography & x <= upper_angiography], upper_angiography), c(0, y_angiography[x >= lower_angiography & x <= upper_angiography], 0), col = "pink")

# Add a legend to the plot
legend("topright", legend = c("Bypass Wait Time", "Angiography Wait Time"), col = c("blue", "red"), fill = c("lightblue", "pink"), lty = 1)

90 % Confidence Interval is narrower for Angiography.

2.

Code

# Set the sample size and number of people who said yes
n_surveyed_adults <- 1031
for_college <- 567

# Calculate the point estimate of the proportion
p_hat <- for_college / n_surveyed_adults
cat("Point estimate for the portion of all adult Americans who believe that a college education is essential for success is: ", p_hat*100,"%", "\n")

Point estimate for the portion of all adult Americans who believe that a college education is essential for success is:  54.99515 % 

Code

# Calculate the standard error
se <- sqrt(p_hat*(1-p_hat)/n_surveyed_adults)

# Calculate the margin of error for a 95% confidence interval
me <- qnorm(0.975)*se

# Calculate the lower and upper bounds of the confidence interval
lower <- p_hat - me
upper <- p_hat + me

# Print the confidence interval
cat("95% Confidence Interval for p: [", lower, ", ", upper, "]", "\n")

95% Confidence Interval for p: [ 0.5195839 ,  0.5803191 ] 

95% Confidence Interval says that, this range of values will have the true population parameter in 95% of the repeated sampling.

3.

Code

me <- 5
population_sd <- (200-30)/4
z_95 <- qnorm(1-0.05/2)

n <- (z_95*population_sd/me)^2
n

[1] 277.5454

4.

A. Assuming that the population mean is 500. Null Hypothesis = There is no difference in the mean income of all senior-level workers and the mean income of senior level female employees.

Code

# Set the significance level
alpha <- 0.05

# Specify the population mean, sample size, sample mean, and sample standard deviation
pop_mean <- 500
n <- 9
sample_mean <- 410
sample_sd <- 90

# Calculate the standard error of the mean
sem <- sample_sd / sqrt(n)

# Calculate the t-statistic
t_stat <- (sample_mean - pop_mean) / sem

# Calculate the degrees of freedom
df <- n - 1

# Calculate the two-tailed p-value
p_value <- 2 * pt(-abs(t_stat), df)

# Determine whether to reject or fail to reject the null hypothesis
if (p_value < alpha) {
  cat("Reject the null hypothesis: the sample mean is significantly different from the population mean.\n")
} else {
  cat("Fail to reject the null hypothesis: there is not sufficient evidence to conclude that the sample mean is different from the population mean.\n")
}

Reject the null hypothesis: the sample mean is significantly different from the population mean.

Code

# Print the test statistics and p-value
cat("t-statistic:", t_stat, "\n")

t-statistic: -3 

Code

cat("p-value:", p_value, "\n")

p-value: 0.01707168 

p value is less than 0.05 and therefore we reject the null hypothesis. Therefore the mean income for all senior-level workers and mean income for senior level female workers are not same. The range of 95% confidence interval is < 500 which means the mean income of the female workers are less than 500 with a 5% significance level.

Code

p_value <- pt(-abs(t_stat), df)
p_value

[1] 0.008535841

It is less than 0.05 we reject null hypothesis.

Code

p_value <- pt(-abs(t_stat), df, lower.tail = F)
p_value

[1] 0.9914642

p value for this one tailed test is 0.4923534 which means we fail to reject null hypothesis since it is greater than 0.05 since we are testing with a 5% significance value. In other words, the true mean of the sample is not greater than population mean

5.

Code

mu <- 500
y_hat_jones <- 519.5
se_jones <- 10
y_hat_smith <- 519.7
se_smith <- 10
n <- 1000

# Set the significance level
alpha <- 0.05

# Specify the population mean, estimated means, and standard errors for each sample
pop_mean <- 500
mean1 <- 519.5
mean2 <- 519.7
se1 <- 10
se2 <- 10
n1 <- 1000
n2 <- 1000

# Calculate the pooled standard error
pooled_se <- sqrt(((n1 - 1) * se1^2 + (n2 - 1) * se2^2) / (n1 + n2 - 2))

# Calculate the t-statistic
t_stat <- (mean1 - mean2 - 0) / (pooled_se * sqrt(1/n1 + 1/n2))

# Calculate the degrees of freedom
df <- n1 + n2 - 2

# Calculate the two-tailed p-value
p_value <- 2 * pt(-abs(t_stat), df)


# Print the test statistics and p-value for both samples
cat("Jones' - t-statistic:", (mean1 - pop_mean) / se1, "p-value:", 2 * pt(-abs((mean1 - pop_mean) / se1), n1 - 1), "\n")

Jones' - t-statistic: 1.95 p-value: 0.05145555 

Code

cat("Smith's - t-statistic:", (mean2 - pop_mean) / se2, "p-value:", 2 * pt(-abs((mean2 - pop_mean) / se2), n2 - 1), "\n")

Smith's - t-statistic: 1.97 p-value: 0.04911426 

6.

Code

# Creating the dataframe
grade_level <- c(rep("6th grade", 100), rep("7th grade", 100), rep("8th grade", 100))
snack <- c(rep("healthy snack", 31), rep("unhealthy snack", 69), rep("healthy snack", 43),
           rep("unhealthy snack", 57), rep("healthy snack", 51), rep("unhealthy snack", 49))
snack_data <- data.frame(grade_level, snack)

We are conducting a Chi-square test in this question since we are testing the association between two categorical variables.

Code

table(snack_data$snack,snack_data$grade_level)

                 
                  6th grade 7th grade 8th grade
  healthy snack          31        43        51
  unhealthy snack        69        57        49

Code

chisq.test(snack_data$snack,snack_data$grade_level,correct = FALSE)

    Pearson's Chi-squared test

data:  snack_data$snack and snack_data$grade_level
X-squared = 8.3383, df = 2, p-value = 0.01547

A p-value smaller than 0.05 indicates that there is a relationship between grade level and the choice of snack.

7.

Code

# Creating the dataframe
Area <- c(rep("Area1", 6), rep("Area2", 6), rep("Area3", 6))
cost <- c(6.2, 9.3, 6.8, 6.1, 6.7, 7.5, 7.5, 8.2, 8.5, 8.2, 7.0, 9.3,
          5.8, 6.4, 5.6, 7.1, 3.0, 3.5)
Area_cost <- data.frame(Area,cost)
one.way <- aov(cost ~ Area, data = Area_cost)
summary(one.way)

            Df Sum Sq Mean Sq F value  Pr(>F)   
Area         2  25.66  12.832   8.176 0.00397 **
Residuals   15  23.54   1.569                   
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The small p-value tells us that the three areas have a difference in means.