One Sample T Test: 3 Example Problems (Humans)?

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In statistics, a is used to test whether or not the mean of a 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:

• H₀: μ = 310 (population mean is equal to 310 pounds)
• H₁: μ ≠ 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 , the p-value associated with t = -3.4817 and degrees of freedom = n-1 = 40-1 = 39 is 0.00149.

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:

• H₀: μ ≤ 82
• H₁: μ > 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 , 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:

• H₀: μ ≥ 10
• H₁: μ < 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 , 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.

The following tutorials provide additional information about hypothesis testing: