Single population hypothesis testing is a statistical method used to test a claim or hypothesis about a population parameter based on sample data. The key components of hypothesis testing include a null hypothesis (\(H_0\)), an alternative hypothesis (\(H_a\)), a test statistic, a critical value or p-value, and a decision rule.
Here are two examples illustrating single population hypothesis testing.
Example 1: Testing a Mean
Scenario: A manufacturer claims that the average lifetime of a particular light bulb is 1,000 hours. A consumer group believes that the true average lifetime is less than 1,000 hours. To test this claim, they take a sample of 30 light bulbs and find that the average lifetime in the sample is 980 hours with a standard deviation of 50 hours.
Step 1: State the Hypotheses
- Null hypothesis (\(H_0\)): \(\mu = 1000\) (The average lifetime of the light bulbs is 1,000 hours.)
- Alternative hypothesis (\(H_a\)): \(\mu < 1000\) (The average lifetime of the light bulbs is less than 1,000 hours.)
Step 2: Choose a Significance Level Choose \( \alpha = 0.05 \).
Step 3: Calculate the Test Statistic Use the formula for the t-test since the population standard deviation is unknown:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{980 - 1000}{50 / \sqrt{30}} \approx -2.19 \]
Where:
- \( \bar{x} \) = sample mean = 980
- \( \mu_0 \) = claimed population mean = 1000
- \( s \) = sample standard deviation = 50
- \( n \) = sample size = 30
Step 4: Determine the Critical Value Using a t-distribution table with \( n-1 = 29 \) degrees of freedom, the critical value for a one-tailed test at \(\alpha = 0.05\) is approximately \(-1.699\).
Step 5: Make a Decision Since \(t = -2.19 < -1.699\), we reject the null hypothesis.
Conclusion: There is sufficient evidence to conclude that the average lifetime of the light bulbs is less than 1,000 hours.
Example 2: Testing a Proportion
Scenario: A poll conducts a survey to determine if the proportion of voters who favor a certain policy is greater than 60%. A random sample of 150 voters shows that 90 of them support the policy.
Step 1: State the Hypotheses
- Null hypothesis (\(H_0\)): \(p = 0.6\) (The proportion of voters in favor of the policy is 60%.)
- Alternative hypothesis (\(H_a\)): \(p > 0.6\) (The proportion of voters in favor of the policy is greater than 60%.)
Step 2: Choose a Significance Level Choose \( \alpha = 0.05 \).
Step 3: Calculate the Test Statistic Using the formula for the z-test for proportions:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} = \frac{0.6 - 0.6}{\sqrt{\frac{0.6(0.4)}{150}}} \approx 1.15 \]
Where:
- \( \hat{p} = \frac{90}{150} = 0.6 \) (sample proportion)
- \( p_0 = 0.6 \) (claimed proportion)
- \( n = 150 \)
Step 4: Determine the Critical Value For a one-tailed test at \(\alpha = 0.05\), the critical z-value is approximately 1.645.
Step 5: Make a Decision Since \(z = 1.15 < 1.645\), we fail to reject the null hypothesis.
Conclusion: There is insufficient evidence to conclude that the proportion of voters who favor the policy is greater than 60%.
These two examples illustrate the hypothesis testing process for both means and proportions in a single population context. Each step involves careful calculations and logical reasoning to make informed decisions based on sample data.