Lecture 37 - Sampling Distributions on the Difference Between Two Means

Overview of Statistical Inference

• Statistical Inference: A statement about an unknown population parameter based on statistics from a sample.
  • Involves estimation (confidence intervals) and statistical tests (hypothesis tests).

Inferences About Two Population Means

• Focus on the difference between two population means, $m<em>1$ and $m</em>2$.

Steps for Making Inferences

1. Identify populations of interest and the unknown parameter ($m<em>1 - m</em>2$).
2. Select random samples from these populations.
3. Collect data from samples.
4. Calculate sample means $\bar{X}<em>1$ and $\bar{X}</em>2$ and find the difference $\bar{X}<em>1 - \bar{X}</em>2$ as the point estimate of $m<em>1 - m</em>2$.

Describing the Sampling Distribution

• A sampling distribution of a statistic is the distribution of values taken by the statistic in numerous simple random samples of the same size from the same population.
• To describe a sampling distribution, consider shape, center, spread, and unusual features.

Assumptions for Sampling Distribution

1. Both samples must be simple random samples and independent.
2. Populations are normally distributed, or if not, sample sizes are sufficiently large (at least 15; if skewed, at least 40).

Sampling Distribution Characteristics

• If population means $m<em>1, m</em>2$ and standard deviations $s<em>1, s</em>2$ are known:
  • Center: $ext{Mean} = m<em>1 - m</em>2$.
  • Spread (Standard Deviation): ext{SD} = rac{ ext{Square Root}(s1^2/n1 + s2^2/n2)} ,
  • Shape: Sampling distribution will be normal under the assumptions.

Situational Examples

Situation 24 - Male vs. Female Students

• Parameter of Interest: Difference in mean IQ of male ( $m<em>1$ ) and female ( $m</em>2$ ) students.
  • Assume $m<em>1 = 102.5$, $s</em>1 = 13.2$.
  • Assume $m<em>2 = 106.8$, $s</em>2 = 12.6$.

Sampling Distributions

1. Sample sizes: $n<em>1 = 28$ (males), $n</em>2 = 33$ (females).
2. Center: $m<em>1 - m</em>2 = 102.5 - 106.8 = -4.3$.
3. Spread: ext{SD} = rac{ ext{Square Root}(13.2^2/28 + 12.6^2/33)} = 3.32
4. Shape: Normal distribution; no unusual features.
   • Thus, $\bar{X}<em>1 - \bar{X}</em>2 hicksim N(-4.3, 3.32)$.

Situation 25 - Student Athletes vs. Non-Athletes

• Parameter of Interest: Difference in mean GPA between student athletes ( $m<em>1$ ) and non-athletes ( $m</em>2$ ).
  • Assume $m<em>1 = 3.23$, $s</em>1 = 0.74$.
  • Assume $m<em>2 = 3.11$, $s</em>2 = 0.82$.

Sampling Distributions

1. Sample sizes: $n<em>1 = 34$ (athletes), $n</em>2 = 102$ (non-athletes).
2. Center: $m<em>1 - m</em>2 = 3.23 - 3.11 = 0.12$.
3. Spread: ext{SD} = rac{ ext{Square Root}(0.74^2/34 + 0.82^2/102)} = 0.1507.
4. Shape: Normal distribution; no unusual features.
   • Thus, $\bar{X}<em>1 - \bar{X}</em>2 hicksim N(0.12, 0.1507)$.

Conducting Probability Assessments

• Using normal distribution properties, probabilities concerning the differences can be derived using Z-scores and normal distribution calculations.
  • Example: For P(ar{X}1 - ar{X}2 < 0), apply normal CDF to find probability values.