t-test in R Programming - Step-by-Step Execution

Concept Flow - t-test

Start

↓

Prepare data samples

↓

Calculate sample means and variances

↓

Compute t-statistic

↓

Calculate degrees of freedom

↓

Find p-value from t-distribution

↓

Compare p-value with significance level

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Decide to reject or not reject null hypothesis

↓

End

The t-test compares the means of two groups by calculating a t-statistic and p-value to decide if the difference is significant.

Execution Sample

R Programming

x <- c(5, 7, 8, 6, 9)
y <- c(10, 12, 11, 14, 13)
t.test(x, y)

This code runs a t-test comparing two small samples x and y to check if their means differ significantly.

Execution Table

Step	Action	Evaluation	Result
1	Prepare sample x	x = c(5,7,8,6,9)	x = [5,7,8,6,9]
2	Prepare sample y	y = c(10,12,11,14,13)	y = [10,12,11,14,13]
3	Calculate mean of x	mean(x)	7
4	Calculate mean of y	mean(y)	12
5	Calculate variance of x	var(x)	2.5
6	Calculate variance of y	var(y)	2.5
7	Calculate sample sizes	length(x), length(y)	5, 5
8	Compute t-statistic	t = (7 - 12) / sqrt(2.5/5 + 2.5/5)	-5
9	Calculate degrees of freedom	df approx	8
10	Calculate p-value	2 * pt(-abs(t), df)	0.0015
11	Compare p-value with 0.05	0.0015 < 0.05	Reject null hypothesis
12	Conclusion	Means differ significantly	End

💡 p-value is less than 0.05, so we reject the null hypothesis that means are equal.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	After Step 10	Final
x	undefined	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]	[5,7,8,6,9]
y	undefined	undefined	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]	[10,12,11,14,13]
mean_x	undefined	undefined	undefined	7	7	7	7	7	7	7	7	7
mean_y	undefined	undefined	undefined	undefined	12	12	12	12	12	12	12	12
var_x	undefined	undefined	undefined	undefined	undefined	2.5	2.5	2.5	2.5	2.5	2.5	2.5
var_y	undefined	undefined	undefined	undefined	undefined	undefined	2.5	2.5	2.5	2.5	2.5	2.5
n_x	undefined	undefined	undefined	undefined	undefined	undefined	undefined	5	5	5	5	5
n_y	undefined	undefined	undefined	undefined	undefined	undefined	undefined	5	5	5	5	5
t_stat	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	-5	-5	-5	-5
df	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	8	8	8
p_value	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	0.0015	0.0015
decision	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	Reject null hypothesis

Key Moments - 3 Insights

Why is the t-statistic negative in the execution table?

Because the mean of x (7) is less than the mean of y (12), the numerator (mean_x - mean_y) is negative, resulting in a negative t-statistic as shown in step 8.

Why do we compare the p-value to 0.05 in step 11?

0.05 is a common significance level threshold; if p-value is less than 0.05, it means the observed difference is unlikely by chance, so we reject the null hypothesis as shown in step 11.

What does degrees of freedom (df) represent in the t-test?

Degrees of freedom approximate the amount of independent information in the data; it affects the shape of the t-distribution used to find the p-value, calculated in step 9.

Visual Quiz - 3 Questions

Look at the execution table at step 8. What is the value of the t-statistic?

A 0

B 5

C -5

D 12

💡 Hint

Check the 'Result' column in step 8 of the execution_table.

At which step does the p-value get calculated?

A Step 7

B Step 10

C Step 5

D Step 12

💡 Hint

Look for the step labeled 'Calculate p-value' in the execution_table.

If the p-value was 0.1 instead of 0.0015, what would be the decision at step 11?

A Accept null hypothesis

B Reject null hypothesis

C Cannot decide

D Run test again

💡 Hint

Compare p-value to 0.05 threshold in step 11 of the execution_table.

Concept Snapshot

t.test(x, y) compares means of two samples.
Calculates t-statistic = difference of means / standard error.
Degrees of freedom affect p-value calculation.
If p-value < 0.05, reject null hypothesis.
Used to check if two groups differ significantly.

Full Transcript

This visual execution traces an R t-test comparing two samples x and y. First, the samples are prepared. Then their means and variances are calculated. Using these, the t-statistic is computed as the difference of means divided by the combined standard error. Degrees of freedom are approximated to determine the shape of the t-distribution. The p-value is found from this distribution. Finally, the p-value is compared to 0.05 to decide if the difference in means is statistically significant. Here, the p-value is very small (0.0015), so the null hypothesis that the means are equal is rejected.

Practice

1.

What does a t-test primarily compare in two groups?

easy

A. The difference between their medians

B. The difference between their averages

C. The total number of observations

D. The variance of each group

Solution

1. Step 1: Understand the purpose of a t-test

   A t-test is used to compare the means (averages) of two groups to see if they are different.

2. Step 2: Identify what is compared

   The test focuses on the average values, not medians, counts, or variances.

3. Final Answer:

   The difference between their averages -> Option B

4. Quick Check:

   t-test compares means = A [OK]

Hint: Remember: t-test checks averages, not medians or counts [OK]

Common Mistakes:

• Confusing mean with median
• Thinking it compares counts
• Assuming it compares variances

2.

Which of the following is the correct way to run a t-test in R comparing two numeric vectors x and y?

easy

A. t_test(x, y)

B. ttest(x, y)

C. t.test(x, y)

D. test.t(x, y)

Solution

1. Step 1: Recall the correct function name in R

   The built-in function for t-test in R is t.test().

2. Step 2: Check the syntax for comparing two vectors

   Using t.test(x, y) compares the two numeric vectors correctly.

3. Final Answer:

   t.test(x, y) -> Option C

4. Quick Check:

   Correct function is t.test() = A [OK]

Hint: Use exact function name t.test() with dot [OK]

Common Mistakes:

• Misspelling the function name
• Using underscores instead of dots
• Confusing function with other test names

3.

What will be the output of this R code snippet?

group1 <- c(5, 7, 8, 6, 9)
group2 <- c(10, 12, 11, 13, 14)
test <- t.test(group1, group2)
round(test$p.value, 3)

medium

A. 0.001

B. 0.05

C. 0.5

D. 1.0

Solution

1. Step 1: Understand the data groups

   Group1 has smaller values around 5-9, Group2 has larger values around 10-14, so means differ.

2. Step 2: Interpret the p-value from t.test()

   Since the groups differ clearly, the p-value will be very small, close to 0.001 after rounding.

3. Final Answer:

   0.001 -> Option A

4. Quick Check:

   Distinct groups give low p-value = B [OK]

Hint: Big difference in groups means very small p-value [OK]

Common Mistakes:

• Assuming p-value is large for different groups
• Confusing p-value with test statistic
• Ignoring rounding in output

4.

Identify the error in this R code for a t-test:

data1 <- c(4, 5, 6)
data2 <- c(7, 8, 9)
test <- t.test(data1 + data2)

medium

A. t.test() cannot compare numeric vectors

B. Missing parentheses in t.test()

C. Vectors data1 and data2 must be data frames

D. Using '+' instead of ',' to separate vectors

Solution

1. Step 1: Check the function call syntax

   The t.test() function expects two separate vectors as arguments separated by a comma.

2. Step 2: Identify the mistake in argument passing

   The code uses '+' to add vectors instead of separating them with a comma, which is incorrect.

3. Final Answer:

   Using '+' instead of ',' to separate vectors -> Option D

4. Quick Check:

   Arguments must be comma-separated = D [OK]

Hint: Separate vectors with comma, not plus sign [OK]

Common Mistakes:

• Adding vectors instead of passing separately
• Confusing function arguments syntax
• Thinking t.test needs data frames

5.

You have two groups of exam scores: before and after a training program. Which t-test is best to check if the training improved scores?

hard

A. Paired t-test using t.test(before, after, paired=TRUE)

B. Two-sample t-test using t.test(before, after)

C. One-sample t-test using t.test(after)

D. Variance test using var.test(before, after)

Solution

1. Step 1: Understand the data relationship

   Scores are from the same students before and after training, so data are paired.

2. Step 2: Choose the correct t-test type

   A paired t-test compares matched pairs to see if scores changed after training.

3. Final Answer:

   Paired t-test using t.test(before, after, paired=TRUE) -> Option A

4. Quick Check:

   Paired data needs paired=TRUE = C [OK]

Hint: Use paired=TRUE for before-after same subjects [OK]

Common Mistakes:

• Using two-sample test for paired data
• Using one-sample test incorrectly
• Confusing variance test with t-test