Confidence Intervals and Hypothesis Testing in Biostatistics
Confidence Intervals
Formulas for Test 2
- Point Estimate (PE) and Margin of Error (MOE)
- Formula:
\text{Point Estimate} + \text{Margin of Error} = \text{PE} + \text{MOE} - Lower Confidence Interval:
\text{Lower C.I.} = \text{PE} - \text{MOE} - Upper Confidence Interval:
\text{Upper C.I.} = \text{PE} + \text{MOE}
- Formula:
### The equations for estimating parameters
- \text{PE} = \text{(mean or proportion)}
- Margin of Error (MOE)
- For proportions:
\text{MOE} = \text{z} \times \sqrt{\frac{p(1-p)}{n}} - For means:
\text{MOE} = t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}
Inference Using Single Sample and Two Samples
Dependent Samples
- Confidence Interval:
- Formula:
\text{C.I.} = d \pm \frac{t{\alpha/2} \cdot Sd}{\sqrt{n}} - Where,
- $d$ = difference of sample means
- $S_d$ = standard deviation of differences
- $n$ = sample size
- Test Statistic:
T = \frac{\bar{x} - d0}{Sd / \sqrt{n}}
- Formula:
Independent Samples
- Non-Pooled is used when variances are unequal
- Test Statistic:
- T = \frac{\bar{x1} - \bar{x2}}{Sp \cdot \sqrt{\frac{1}{n1} + \frac{1}{n_2}}}
- Where:
- Sp = \sqrt{\frac{(n1 - 1)S1^2 + (n2 - 1)S2^2}{n1 + n_2 - 2}}
Bartlett’s F Test for Variances
- Formula:
- F = \frac{S1^2}{S2^2} where $S1^2 > S2^2$
- Degrees of freedom
- df{numerator} = n{1} - 1
- df{denominator} = n{2} - 1
Quick TI cheats
- Confidence Interval:
- Use STAT
- Single Sample Tests:
- For mean: TINTERVAL for confidence interval
- For proportion: 1-PROPZINT
- Two Independent Samples:
- Pooled: 2-SAMPTTEST
- Non-Pooled: 2-SAMPTTEST
Principles of Biostatistics - Test 2 Instructions
- Show all work for credit.
- Using calculators:
- Allowed for checking work; must show commands and round answers to 4 decimal places.
Hypothesis Testing Example: Comparing Average Salaries
Department A vs. Department B
- Sample Size from A:
- n_A = 40
- Average Salary from A:
- \bar{x}_A = 55000
- Standard Deviation from A:
- S_A = 5000
- Sample Size from B:
- n_B = 35
- Average Salary from B:
- \bar{x}_B = 53500
- Standard Deviation from B:
- S_B = 4800
Hypotheses
- State the hypotheses:
- Null hypothesis:
- H0: \sigma{A}^2 = \sigma_{B}^2
- Alternative hypothesis:
- HA: \sigma{A}^2 > \sigma_{B}^2
Test Statistic
- Calculate the test statistic using the F-test formula:
- F = \frac{SA^2}{SB^2}
Decision and Interpretation
- Determine p-value from the F-distribution.
- Decision:
- Reject $H_0$
- Fail to reject $H_0$
Orange Yield Comparison Example: Florida vs. California
- Sample Means from Table 1:
- Florida Mean: \bar{x}_{FL} = 31.67
- California Mean: \bar{x}_{CA} = 28.73
- Hypothesis Test:
- Test if yield in FL is more than in CA.
- Statistical outputs:
- p-value from Bartlett's Test:
- p = 0.179
Hypotheses for Yield Comparison
- State the hypotheses:
- Null: H0: \mu{FL} \leq \mu_{CA}
- Alternative: HA: \mu{FL} > \mu_{CA}
Interpretation of Results
- Decision:
- If p > 0.05, fail to reject $H_0$
- If p \leq 0.05, reject $H_0$
Daily Step Count Analysis from Fitness Study
Study Details
- Sample Size: 16 individuals
- Mean Daily Step Count: 779.38 mg/dL
Data Input
- Daily Step Count Before and After table.
- Analyze possible shifts in average step count.
Confidence Interval Calculation
- Construct a 95% confidence interval:
- Compute t_{\frac{\alpha}{2}} & proposals for CI
- Results: Calculate and round as necessary.
Evaluation of New Study Results
- Findings suggest:
- Confidence Interval for weight loss:
- Reported as (0.51, 0.69)
- Calculate:
- Point Estimate (P.E.)
- Margin of Error (M.O.E.)
- \text{P.E.} = \frac{0.51 + 0.69}{2} = 0.601
- \text{M.O.E.} = \frac{0.69 - 0.51}{2} = 0.09
Interpretation of the Results
Interpretation of Confidence Interval:
- We are 90% confident that the true, unknown population mean of weight loss μ is within the interval (0.51, 0.69).
- The results suggest a reasonable expectation about weight loss under the new program
Overall conclusion of tests must adhere to statistical significance criteria, coupled with clear articulation of conditions and methodology utilized during analyses.
Hypothesis Testing Example: Comparing Average Salaries
Department A vs. Department B
- Sample Size from A:
n_A = 40 - Average Salary from A:
\bar{x}_A = 55000 - Standard Deviation from A:
S_A = 5000 - Sample Size from B:
n_B = 35 - Average Salary from B:
\bar{x}_B = 53500 - Standard Deviation from B:
S_B = 4800
Hypotheses
- State the hypotheses:
- Null hypothesis:
H0: \sigmaA^2 = \sigma_B^2 - Alternative hypothesis:
HA: \sigmaA^2 > \sigma_B^2
- Null hypothesis:
Test Statistic
- Calculate the test statistic using the F-test formula:
- Formula:
F = \frac{SA^2}{SB^2} - Substitute the values:
F = \frac{(5000)^2}{(4800)^2} - Calculation step:
F = \frac{25000000}{23040000} \approx 1.087
- Formula:
Decision and Interpretation
- Determine the p-value from the F-distribution based on the calculated F statistic and degrees of freedom.
- Decision:
- If p-value < significance level (e.g., 0.05), reject $H_0$.
- If p-value ≥ significance level, fail to reject $H_0$.
Orange Yield Comparison Example: Florida vs. California
- Sample Means from Table 1:
- Florida Mean:
\bar{x}_{FL} = 31.67 - California Mean:
\bar{x}_{CA} = 28.73 - Hypothesis Test:
- Test if yield in FL is more than in CA.
Hypotheses for Yield Comparison
- State the hypotheses:
- Null:
H0: \mu{FL} \leq \mu_{CA} - Alternative:
HA: \mu{FL} > \mu_{CA}
- Null:
Interpretation of Results
- Decision:
- If p > 0.05, fail to reject $H_0$
- If p \leq 0.05, reject $H_0$
Daily Step Count Analysis from Fitness Study
Study Details
- Sample Size: 16 individuals
- Mean Daily Step Count: 779.38 mg/dL
Data Input
- Daily Step Count Before and After table.
- Analyze possible shifts in average step count.
Confidence Interval Calculation
- Construct a 95% confidence interval:
- Compute t_{\frac{\alpha}{2}} & proposals for CI.
- Results: Calculate and round as necessary.
Evaluation of New Study Results
- Findings suggest:
- Confidence Interval for weight loss:
- Reported as (0.51, 0.69)
- Calculate:
- Point Estimate (P.E.)
- Margin of Error (M.O.E.)
- \text{P.E.} = \frac{0.51 + 0.69}{2} = 0.601
- \text{M.O.E.} = \frac{0.69 - 0.51}{2} = 0.09
Interpretation of the Results
- Interpretation of Confidence Interval:
- We are 90% confident that the true, unknown population mean of weight loss μ is within the interval (0.51, 0.69).
- The results suggest a reasonable expectation about weight loss under the new program.
- Overall conclusion of tests must adhere to statistical significance criteria, coupled with clear articulation of conditions and methodology utilized during analyses.