Hypothesis Testing Study Notes

Example problem using p-tests.
Anticipation of a worksheet related to tests will be handed out on Thursday or posted online.

Example data for two sample t-tests:
- Sample 1:
- Mean (x̄₁) = 8.1
- Standard Deviation (Sₓ₁) = 2.3
- Sample Size (n₁) = 8
- Sample 2:
- Mean (x̄₂) = 9.4
- Standard Deviation (Sₓ₂) = 2.8
- Sample Size (n₂) = 10
Discussion on the type of t-test:
- Generally use the "not equals" hypothesis most of the time unless more information is provided.
t-Test Statistic Calculation:
- Result: t-test statistic = -1.08
- p-value = 0.29
Interpretation of p-value:
- If the p-value > alpha: Fail to reject the null hypothesis.
- Claim cannot be deemed true, only evidence to reject or fail to reject the null hypothesis.
Discussion on beta risk and its implications on the p-value calculation.

Restate the claim:
- Example Proportion Claim: Population proportion (p) is equal to some value, often denoted as p₀.
Significance Level (alpha): Must be specified for hypothesis testing.
Z Test Statistic for Proportion:
- Formula: $Z = \frac{\hat{p} - p<em>0}{\sqrt{\frac{p</em>0(1 - p_0)}{n}}}$
- Where (\hat{p}) is the sample proportion, (p_0) is the hypothesized population proportion.
Using the One Proportion Z-Test on calculators:
- Instructions: Access the test under the "stat" menu and follow prompts for input.

Claim: 75% of union members support basic demands.
Company’s representative believes it's lower, and they use a significance level of 10%.
- Sample Data:
- 87 out of 125 sampled support the demands.
Calculation of sample proportion (p̂):
- Resulting p̂ from data gathered = 0.696 (which is 87/125).
Input Data for One Proportion Z-Test:
- Claim (p₀) = 0.75 (input as decimal, not percentage).
- Alternative hypothesis: p < 0.75.
Resulting Z Test Statistic and p-value:
- Z Test Statistic = -1.39
- p-value = 0.081.
Final interpretation of the p-value in relation to alpha (0.10):
- Since 0.081 < 0.10, there is evidence to support the claim that less than 75% support the demands.

Homework Assignment: Related to Proportions, Chapter 6.1 only (excluding Chapter 6.3).
Next Topics: Moving into the Chi-Square tests, particularly focusing on independence tests.

Focus: Determining relationships between categorical variables.
Problem scenario: A beef distributor explores the relationship between geographic region and preferred meat cuts.
Sample size includes 500 consumers (300 North, 200 South):
- Preferences:
- Cut A: 150
- Cut B: 275
- Cut C: 75
Observation and expected tables to be constructed.

Null Hypothesis (H₀): No relationship between geographic region and cuts of meat.
Alternative Hypothesis (H₁): There is a relationship between geographic region and cuts of meat.
Expected Table Formation:
- Expected counts calculated under null hypothesis. Logic behind each distribution of observed vs. expected counts.

Formula for Chi-Square: $\chi^2 = \sum \frac{(O - E)^2}{E}$
- Where O = observed values and E = expected values.
Importance of summing across all cells to get the chi-square statistic.

Chi-square distribution is only right-sided (values > 0).
Critical values determined based on degrees of freedom and significance level.
- Example: For 1% significance level with degrees of freedom, check the corresponding chi-square critical value from the chi-square table.
Explanation on how degrees of freedom are calculated:
- Formula: $df = (r - 1)(c - 1)$ where r = number of rows, c = number of columns.
- Importance of understanding degrees of freedom in chi-square table interpretation.

Importance of sample sizes greater than 5 in expected counts.
Large chi-square value suggests evidence against the null hypothesis, while small chi-square values do not.
Expected count calculations using:
$E = \frac{(row\ total) \times (column\ total)}{(total\ sample\ size)}$
Suggested assignment: Two chi-square independence problems and two population problems.

Anticipation of further practice and analysis on chi-square tests in upcoming classes.