Table of Contents
A one sample t-test is a statistical test used to determine if the mean of a single sample is significantly different from a known or hypothesized population mean. It is commonly used in research and data analysis to draw conclusions about a population based on a smaller sample. Three examples of one sample t-tests include:
1. Testing the effectiveness of a new medication: In this scenario, a researcher may collect data from a sample of patients who have been given the new medication and then use a one sample t-test to compare the mean effectiveness of the medication with a known population mean.
2. Examining the impact of a new advertising campaign: A company may conduct a one sample t-test to determine if the mean sales after implementing a new advertising campaign is significantly different from the mean sales before the campaign was launched.
3. Investigating the average income of a specific population: A government agency may use a one sample t-test to compare the mean income of a particular demographic group with a known population mean, in order to identify any significant differences in income levels.
One Sample T Test: 3 Example Problems
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.
Additional Resources
The following tutorials provide additional information about hypothesis testing: