Statistics Hypothesis Testing and Confidence Intervals Notes

  • True/False Statements (1-5)

    • Statement 1: True. For a two-sided hypothesis test, if 1 + CL = α\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 (xˉ\bar{x}) = 149.43
    • Sample standard deviation (s) = 53.30
    • Intamin:
    • n = 14
    • xˉ\bar{x} = 179.63
    • s = 103.30
    • Vekoma:
    • n = 33
    • xˉ\bar{x} = 87.90
    • s = 32.50
    • Arrow:
    • n = 23
    • xˉ\bar{x} = 93.48
    • s = 49.89

  • Constructing Confidence Intervals (CI)

    • 90% CI for B&M:
    • Formula: xˉ±z∗⋅sn\bar{x} \pm z^* \cdot \frac{s}{\sqrt{n}}
    • Critical value ($z^*$) for 90% = 1.645
    • CI calculation:
      • 149.43±1.645⋅53.3046149.43 \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: xˉ±t∗⋅sn\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: H0:μ=95H_0: \mu = 95.
    • Compute test statistic t:
      • Formula: t=xˉ−μs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}
      • t=87.90−9532.50/33t = \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: p^=xn=84140=0.60\hat{p} = \frac{x}{n} = \frac{84}{140} = 0.60
    • Wald Interval Requirements:
    • Condition 1: np≥5np \geq 5 and n(1−p)≥5n(1-p) \geq 5.
    • Condition Check:
      • 140⋅0.60=84≥5140\cdot0.60=84 \geq 5 and 140⋅0.40=56≥5140\cdot0.40=56 \geq 5
    • 95% Confidence Interval for Proportion:
    • Formula: p^±z∗p^(1−p^)n\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
    • CI Calculation:
      • 0.60±1.960.60⋅0.401400.60 \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:
    • p^=48150=0.32\hat{p} = \frac{48}{150} = 0.32
    • Hypothesis Setup:
    • $H_0: p = 0.35$
    • Ha:p≠0.35H_a: p \neq 0.35
    • Test Statistic Calculation:
    • z=p^−p<em>0p</em>0(1−p0)nz = \frac{\hat{p} - p<em>0}{\sqrt{\frac{p</em>0(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∗≈2.33z^* \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.