Confidence intervals in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Calculate 95% confidence interval for mean

What is the output of this R code that calculates a 95% confidence interval for the mean of a numeric vector?

R Programming

data <- c(5, 7, 8, 6, 9)
mean_val <- mean(data)
std_err <- sd(data) / sqrt(length(data))
conf_int <- mean_val + c(-1, 1) * qt(0.975, df=length(data)-1) * std_err
round(conf_int, 2)

A [4.50, 9.50]

B [5.00, 9.00]

C [5.04, 8.96]

D [6.00, 8.00]

💡 Hint

Remember to use the t-distribution quantile for the confidence interval with small samples.

The code calculates the mean and standard error, then uses the t-distribution quantile for 95% confidence with 4 degrees of freedom. The resulting interval is [5.04, 8.96].

🧠 Conceptual

intermediate

Interpretation of confidence interval

Which statement correctly describes the meaning of a 95% confidence interval for a population mean?

A There is a 95% chance the population mean lies within the interval calculated from the sample.

B 95% of the sample data points lie within the interval.

C The population mean is exactly at the midpoint of the interval.

D If we repeat the sampling many times, 95% of the calculated intervals will contain the true population mean.

💡 Hint

Think about what happens if you take many samples and calculate intervals each time.

A 95% confidence interval means that if we repeat the sampling process many times, about 95% of those intervals will contain the true population mean. It does not mean the probability for a single interval.

❓ Predict Output

advanced

Confidence interval for proportion

What is the output of this R code that calculates a 90% confidence interval for a proportion using the normal approximation?

R Programming

x <- 45
n <- 100
p_hat <- x / n
z <- qnorm(0.95)
se <- sqrt(p_hat * (1 - p_hat) / n)
conf_int <- p_hat + c(-1, 1) * z * se
round(conf_int, 3)

A [0.368, 0.532]

B [0.400, 0.500]

C [0.350, 0.550]

D [0.420, 0.480]

💡 Hint

Use the z-score for 90% confidence and calculate standard error for proportion.

The sample proportion is 0.45. The z-score for 90% confidence is about 1.645. The standard error is sqrt(p*(1-p)/n). The interval is [0.368, 0.532].

🔧 Debug

advanced

Identify error in confidence interval code

What error does this R code produce when calculating a 95% confidence interval for the mean?

R Programming

data <- c(10, 12, 14, 16, 18)
mean_val <- mean(data)
std_err <- sd(data) / length(data)
conf_int <- mean_val + c(-1, 1) * qt(0.975, df=length(data)-1) * std_err
conf_int

A The interval is too narrow because standard error is calculated incorrectly.

B SyntaxError due to missing parentheses.

C Runtime error: division by zero.

D The code runs correctly and outputs the confidence interval.

💡 Hint

Check the formula for standard error of the mean.

Standard error should be sd(data) divided by the square root of sample size, not by the sample size itself. This makes the interval too narrow and misleading.

🚀 Application

expert

Calculate and interpret confidence interval for difference of means

Given two independent samples, what is the 95% confidence interval for the difference of their means? Use this R code and select the correct output.

R Programming

group1 <- c(20, 22, 19, 24, 21)
group2 <- c(18, 17, 20, 16, 19)
mean_diff <- mean(group1) - mean(group2)
se_diff <- sqrt(var(group1)/length(group1) + var(group2)/length(group2))
t_val <- qt(0.975, df=min(length(group1), length(group2)) - 1)
conf_int <- mean_diff + c(-1, 1) * t_val * se_diff
round(conf_int, 2)

A [1.00, 5.00]

B [0.11, 6.29]

C [-1.50, 4.50]

D [0.00, 7.00]

💡 Hint

Calculate difference of means, standard error of difference, then use t-value for confidence interval.

The difference of means is 21.2 - 18 = 3.2. The standard error of difference is about 1.11. Using t-value ~2.776 for df=4, the interval is [0.11, 6.29].