Statistics Hypothesis Testing and Confidence Intervals Notes
True/False Statements (1-5)
- Statement 1: True. For a two-sided hypothesis test, if 1 + CL = $\alpha$, the conclusions for the test and the associated confidence interval will agree.
- Statement 2: True. The margin of error is composed of the critical value and standard error.
- Statement 3: True. A test statistic gives the distance between the sample statistic and the null value.
- Statement 4: True. Increasing the level of confidence increases the margin of error for a confidence interval.
- Statement 5: False. In a one-sided test, only one rejection region is created.
Summary Statistics for Roller Coasters (Manufacturer Data)
- B&M:
- Sample size (n) = 46
- Sample mean height ($\bar{x}$) = 149.43
- Sample standard deviation (s) = 53.30
- Intamin:
- n = 14
- $\bar{x}$ = 179.63
- s = 103.30
- Vekoma:
- n = 33
- $\bar{x}$ = 87.90
- s = 32.50
- Arrow:
- n = 23
- $\bar{x}$ = 93.48
- s = 49.89
Constructing Confidence Intervals (CI)
- 90% CI for B&M:
- Formula: $\bar{x} \pm z^* \cdot \frac{s}{\sqrt{n}}$
- Critical value ($z^*$) for 90% = 1.645
- CI calculation:
- $149.43 \pm 1.645 \cdot \frac{53.30}{\sqrt{46}}$
- Keep at least 2 decimals throughout.
- 95% CI for Arrow:
- Use $t^*$ value for small samples (n=23).
- Critical value for 95% and df = n-1 = 22:
- CI calculation: $\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}$ and keep at least 2 decimals throughout.
- Test Statistic for Vekoma Hypothesis Testing:
- Hypothesize mean height: $H_0: \mu = 95$.
- Compute test statistic t:
- Formula: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$
- $t = \frac{87.90 - 95}{32.50 / \sqrt{33}}$, keep at least 2 decimals.
Confidence Interval for Mean Height Difference:
- CI: [41.34, 81.71]
- Conclusion for t-test: Evidence suggests one mean (B&M) is larger than the other (Vekoma) since the interval does not contain 0.
Hypothesis Testing SC Voters
- Point Estimate for Candidate A:
- Formula: $\hat{p} = \frac{x}{n} = \frac{84}{140} = 0.60$
- Wald Interval Requirements:
- Condition 1: $np \geq 5$ and $n(1-p) \geq 5$.
- Condition Check:
- $140\cdot0.60=84 \geq 5$ and $140\cdot0.40=56 \geq 5$
- 95% Confidence Interval for Proportion:
- Formula: $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
- CI Calculation:
- $0.60 \pm 1.96 \sqrt{\frac{0.60 \cdot 0.40}{140}}$, keep at least 2 decimals throughout.
- Interpretation: The confidence interval estimates the true proportion of voters who support candidate A.
Columbia Residents Theater Visit Hypothesis Testing:
- Point Estimate for p:
- $\hat{p} = \frac{48}{150} = 0.32$
- Hypothesis Setup:
- $H_0: p = 0.35$
- $H_a: p \neq 0.35$
- Test Statistic Calculation:
- $z = \frac{\hat{p} - p0}{\sqrt{\frac{p0(1-p_0)}{n}}}$
- Compute with $p_0 = 0.35$
- p-value Reporting:
- Calculate using standard z-table for computed z.
- Conclusion: Compare p-value with significance level (0.05).
Critical Value and p-value Reporting:
- Critical Value for 98% CI:
- Use z-table, $z^* \approx 2.33$ for n = 28 (df = 27).
- p-value Reporting:
- For $H_0: p = 0.75$ and $z = 1.75$, calculate using standard normal distribution table.