# One Sample T Test: 3 Example Problems

In statistics, a one sample t-test is used to test whether or not the mean of a population is equal to some value.

The following examples show how to perform the three types of one sample t-tests:

• Two-tailed one sample t-test
• Right-tailed one sample t-test
• Left-tailed one sample t-test

Let’s jump in!

## Example 1: Two-Tailed One Sample T-Test

Suppose we want to know whether or not the mean weight of a certain species of turtle is equal to 310 pounds.

To test this, will perform a one-sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles with the following information:

• Sample size n = 40
• Sample mean weight x = 300
• Sample standard deviation s = 18.5

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H0μ = 310 (population mean is equal to 310 pounds)
• H1μ ≠ 310 (population mean is not equal to 310 pounds)

Step 3: Calculate the test statistic t.

t = (x – μ) / (s/√n) = (300-310) / (18.5/√40) = -3.4187

Step 4: Calculate the p-value of the test statistic t.

According to the T Score to P Value Calculator, the p-value associated with t = -3.4817 and degrees of freedom = n-1 = 40-1 = 39 is 0.00149.

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean weight of this species of turtle is not equal to 310 pounds.

## Example 2: Right-Tailed One Sample T-Test

Suppose we suspect that the mean exam score on a certain college entrance exam is greater than the accepted mean score of 82.

To test this, will perform a right-tailed one-sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of exam scores with the following information:

• Sample size n = 60
• Sample mean x = 84
• Sample standard deviation s = 8.1

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H0μ ≤ 82
• H1μ > 82

Step 3: Calculate the test statistic t.

t = (x – μ) / (s/√n) = (84-82) / (8.1/√60) = 1.9125

Step 4: Calculate the p-value of the test statistic t.

According to the T Score to P Value Calculator, the p-value associated with t = 1.9125 and degrees of freedom = n-1 = 60-1 = 59 is 0.0303.

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean exam score on this particular exam is greater than 82.

## Example 3: Left-Tailed One Sample T-Test

Suppose we suspect that the mean height of a particular species of plant is less than the accepted mean height of 10 inches.

To test this, will perform a left-tailed one-sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of plants with the following information:

• Sample size n = 25
• Sample mean x = 9.5
• Sample standard deviation s = 3.5

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H0μ ≥ 10
• H1μ < 10

Step 3: Calculate the test statistic t.

t = (x – μ) / (s/√n) = (9.5-10) / (3.5/√25) = -0.7143

Step 4: Calculate the p-value of the test statistic t.

According to the T Score to P Value Calculator, the p-value associated with t = -0.7143 and degrees of freedom = n-1 = 25-1 = 24 is 0.24097.

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean height for this particular plant species is less than 10 inches.