Survey Sampling and Inference: Hypothesis Testing for Population Means (Chapter 7 Review)

Survey Sampling and Inference

Hypothesis Testing for Population Means

This section reviews the process of hypothesis testing for population means using sample data, focusing on the application of the one-sample t-test in a real-world context.

Formulating Hypotheses

• Null Hypothesis (H0): The null hypothesis states that there is no change or effect. In this scenario, it asserts that the mean years of experience among nursing staff is still 12 years.

• Alternative Hypothesis (HA): The alternative hypothesis posits that the mean years of experience has increased compared to 12 years.

Statistical Notation:

• $H_0: \mu = 12$

• $H_A: \mu > 12$

Example: Testing whether the average years of experience for nurses in 2024 is greater than the previously reported average of 12 years.

Appropriateness of the One-Sample t-Test

• One-sample t-test: Used to compare the mean of a single sample to a known value (population mean) when the population standard deviation is unknown.

• Conditions:

  • Sample is randomly selected.

  • Population standard deviation is unknown.

  • Sample size is moderate (n = 30), which is generally sufficient for the Central Limit Theorem to apply.

• Conclusion: A one-sample t-test is appropriate because the sample mean is being compared to a known population mean, and the sample standard deviation is used as an estimate.

Example: Comparing the mean years of experience from a sample of 30 nurses to the historical mean of 12 years.

Calculating the t-Test Statistic

• Formula:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

• $\bar{x}$ = sample mean = 15

• $\mu_0$ = population mean under H0 = 12

• $s$ = sample standard deviation = 10

• $n$ = sample size = 30

Calculation:

$t = \frac{15 - 12}{10 / \sqrt{30}} = \frac{3}{10 / 5.477} = \frac{3}{1.8257} \approx 1.644$

Rounded to three decimal places: $t \approx 1.644$

Interpreting Results and Drawing Conclusions

• Critical t-value: At a 0.05 significance level with 29 degrees of freedom, the critical value is 2.045.

• p-value: The p-value associated with the calculated t-statistic is 0.0181.

• Decision Rule:

  • If the p-value is less than the significance level ($\alpha = 0.05$), reject the null hypothesis.

  • If the t-statistic exceeds the critical value, also reject the null hypothesis.

• Conclusion in Context: Since the p-value (0.0181) is less than 0.05, there is sufficient evidence to conclude that the mean years of experience among the hospital's nursing staff has increased compared to 12 years.

Example: The nurse manager can report that, based on the sample, the average years of experience for nurses is significantly higher than the previous average.

Summary Table: Hypothesis Testing Steps

Step	Description	Application to Example
1. State Hypotheses	Formulate null and alternative hypotheses	$H_0: \mu = 12$, $H_A: \mu > 12$
2. Select Test	Choose appropriate statistical test	One-sample t-test
3. Calculate Statistic	Compute t-statistic using sample data	$t \approx 1.644$
4. Compare to Critical Value / p-value	Assess significance	Critical value = 2.045; p-value = 0.0181
5. Draw Conclusion	Interpret results in context	Mean years of experience has increased

Additional info: The one-sample t-test is robust to moderate deviations from normality, especially with sample sizes of 30 or more.