Hypothesis Testing: Two Sample Tests

Hypothesis Testing with Two Samples

Null and Alternative Hypotheses

• Null Hypothesis: No significant difference between the sample means of the two groups.
• Alternative Hypothesis: There is a significant difference between the sample means of the two groups.

Critical Region

• The critical region is determined by the alpha level $\alpha$. For $\alpha = 0.05$ in a two-tailed test, the critical values are $\pm 1.96$.

Formulas for Test Statistic

The test statistic $z$ is calculated differently for two-sample tests. The general formula is:

$z = \frac{\bar{x}<em>1 - \bar{x}</em>2 - (\mu<em>1 - \mu</em>2)}{\text{Standard Deviation of Sampling Distribution}}$

Where:

• $\bar{x}<em>1$ and $\bar{x}</em>2$ are the sample means of the two groups.
• $\mu<em>1$ and $\mu</em>2$ are the population means of the two groups.

Under the null hypothesis, $\mu<em>1 - \mu</em>2 = 0$, so the formula simplifies to:

$z = \frac{\bar{x}<em>1 - \bar{x}</em>2}{\text{Standard Deviation of Sampling Distribution}}$

Expanding the Formula

The standard deviation of the sampling distribution needs to be calculated. The formula depends on whether the population standard deviations are known.

• If population standard deviations are known:

  $\text{Standard Deviation of Sampling Distribution} = \sqrt{\frac{\sigma<em>1^2}{n</em>1} + \frac{\sigma<em>2^2}{n</em>2}}$

  Where:

  • $\sigma<em>1^2$ and $\sigma</em>2^2$ are the variances of the two populations.
  • $n<em>1$ and $n</em>2$ are the sample sizes of the two groups.

• If population standard deviations are unknown:

  $\text{Standard Deviation of Sampling Distribution} = \sqrt{\frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}}$

  Where:

  • $s<em>1^2$ and $s</em>2^2$ are the sample variances of the two groups.
  • $n<em>1$ and $n</em>2$ are the sample sizes of the two groups.

Decision Making

• After calculating the test statistic $z$, compare it to the critical values.
• If |z| > 1.96 (for $\alpha = 0.05$), reject the null hypothesis.
• If $|z| \leq 1.96$, fail to reject the null hypothesis.

Factors Influencing the Decision

1. Size of the Difference Between Sample Statistics:

   • If the difference between $\bar{x}<em>1$ and $\bar{x}</em>2$ is large, it is more likely to reject the null hypothesis. This means the calculated $z$ value will likely fall beyond the critical value.
   • Example: If group 1 scores 90% and group 2 scores 70%, the large difference increases the likelihood of rejecting the null hypothesis.

2. Alpha Level:

   • A larger alpha level (e.g., $\alpha = 0.10$) makes it more likely to reject the null hypothesis.
   • Analogy: A lenient teacher (higher alpha) is more likely to give higher grades, making it easier to score well.

3. One-Tailed vs. Two-Tailed Test:

   • One-tailed tests are more likely to reject the null hypothesis because the entire alpha level is concentrated in one direction.
   • Two-tailed tests split the rejection region, making it more strict.

4. Sample Size:

   • A larger sample size makes it more likely to reject the null hypothesis.
   • A larger sample size makes the denominator of the $z$ statistic smaller, resulting in a larger $z$ value if the numerator (difference in means) is constant.
   • If $n$ is large and the denominator is small, then $z$ is large.
   • A survey of 1000 people is more reliable than a survey of 10 people.

Example Question

• Question: Do athletes in different sports (basketball vs. football) vary in terms of their readiness for college, based on college entrance exam scores?

Solving the Example

1. Problem Statement: Determine if there is a significant difference in the average entrance exam scores of basketball and football players.

2. Test Selection: Use an independent two-sample z-test.

   • The exam scores of football players do not impact the exam scores of basketball players, hence they are independent.
   • Sample sizes are greater than 30.

3. Critical Values:

   • Given $\alpha = 0.05$ in a two-tailed test, the critical values are $\pm 1.96$.

4. Formula:

   $z = \frac{\bar{x}<em>1 - \bar{x}</em>2}{\sqrt{\frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}}}$

5. Calculations:

   • Given:

     • Basketball players: $\bar{x}<em>1 = 460$, $s</em>1 = 92$, $n_1 = 102$
     • Football players: $\bar{x}<em>2 = 442$, $s</em>2 = 57$, $n_2 = 117$

   • Standard Deviation of Sampling Distribution:

     $\sqrt{\frac{92^2}{102} + \frac{57^2}{117}} = \sqrt{\frac{8464}{102} + \frac{3249}{117}} = \sqrt{83 + 27.77} = 10.52376$

   • Z-score:

     $z = \frac{460 - 442}{10.52376} = \frac{18}{10.52376} = 1.71$

6. Decision:

   • Since $z = 1.71$ is between $-1.96$ and $+1.96$, fail to reject the null hypothesis.

7. Conclusion:

   • There is no evidence to suggest that basketball and football players have significantly different college entrance exam scores. The observed difference is likely due to random chance.