WK10: Statistical Inference - Comparing Two Means: Test of Significance for Two Independent Populations

Test of Significance for Two Independent Populations

Steps

  • Step 1: State the null and alternative hypotheses, and the level of significance α\alpha.

    • Parameter of interest: Difference between means of two populations.
    • Null hypothesis: H<em>0:μ</em>1μ2=0H<em>0: \mu</em>1 - \mu_2 = 0 (no difference in means).
    • Alternative hypothesis:
      • Two-sided: H<em>a:μ</em>1μ20H<em>a: \mu</em>1 - \mu_2 \neq 0
      • One-sided: H<em>a:μ</em>1μ<em>2<0H<em>a: \mu</em>1 - \mu<em>2 < 0 or Ha: \mu1 - \mu2 > 0
  • Step 2: Check conditions and calculate the test statistic.

    • Conditions:
      • Both samples from the populations of interest must be random.
      • Population distributions are either close to normal or sample sizes are large.
    • Test statistic:
      • General formula: point estimatenull valuestandard error\frac{\text{point estimate} - \text{null value}}{\text{standard error}}
      • For two independent samples: t=xˉ<em>1xˉ</em>2(μ<em>1μ</em>2)s<em>12n</em>1+s<em>22n</em>2t = \frac{\bar{x}<em>1 - \bar{x}</em>2 - (\mu<em>1 - \mu</em>2)}{\sqrt{\frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}}}
      • Under the null hypothesis (μ<em>1μ</em>2=0\mu<em>1 - \mu</em>2 = 0): t=xˉ<em>1xˉ</em>2s<em>12n</em>1+s<em>22n</em>2t = \frac{\bar{x}<em>1 - \bar{x}</em>2}{\sqrt{\frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}}}
  • Step 3: Find the P-value.

    • The test statistic follows a T distribution.
    • Degrees of freedom: Use the conservative approach: df=min(n<em>11,n</em>21)df = \min(n<em>1 - 1, n</em>2 - 1)
    • Use Table C to find the P-value range based on the t statistic and degrees of freedom.
  • Step 4: Make a statistical decision and draw a conclusion.

    • If P-value < \alpha, reject the null hypothesis.
      • The samples provide statistically significant evidence that the alternative hypothesis is true at level of significance α\alpha.
    • If P-value > \alpha, fail to reject the null hypothesis.
      • The samples do not provide enough evidence to support the alternative hypothesis at level of significance α\alpha.

Example: Exercise and Pulse Rate

  • Study: Compare the mean resting pulse rate of adults who regularly exercise to those who do not.
  • Question: Do these two populations differ in their mean resting pulse rates?
  • Level of Significance: α=0.05\alpha = 0.05
  • Direction of Difference: Non-exercisers - Exercisers
Hypotheses
  • Null hypothesis: H<em>0:μ</em>non-exercisersμexercisers=0H<em>0: \mu</em>{\text{non-exercisers}} - \mu_{\text{exercisers}} = 0
    • The mean resting pulse rate of adults who do not regularly exercise is the same as that of those who do.
  • Alternative hypothesis: H<em>a:μ</em>non-exercisersμexercisers0H<em>a: \mu</em>{\text{non-exercisers}} - \mu_{\text{exercisers}} \neq 0
    • The mean resting pulse rate of adults who do not regularly exercise is different from that of those who do.
Conditions
  • The sum of sample sizes is greater than 40 (specifically, 60), so proceed with T procedures.
Test Statistic
  • Given sample means and standard deviations, the calculated test statistic is t=3.96t = 3.96
  • Degrees of freedom: df=min(n<em>11,n</em>21)=28df = \min(n<em>1 - 1, n</em>2 - 1) = 28
P-value
  • For df=28df = 28, t=3.96t = 3.96 is larger than the last tabulated value of 3.674.
  • Therefore, the P-value is less than 0.001 (for a two-sided test).
Conclusion
  • Since the P-value (< 0.001) is less than the level of significance (α=0.05\alpha = 0.05), we reject the null hypothesis.
  • The samples provide statistically significant evidence that, at a 5% level of significance, the mean resting pulse rate of adults who do not regularly exercise is different from that of adults who do regularly exercise.