Biostatistics Notes and Confidence Intervals

Biostatistics Study Notes

Example 6.3 - Estimating Proportion of Adults Who Drink Occasionally

  • Research Question: Estimate the proportion of adults who drink beer, wine, or hard liquor at least occasionally.
  • Sample Size: 40 adults selected randomly from a large city.
  • Responses:
    • 23 out of 40 responded "Yes";
    • Data breakdown: Yes, Yes, Yes, No, Yes, No, No, No, Yes, Yes, No, Yes, No, Yes, No, No, Yes, No, No, Yes, Yes, Yes, Yes, Yes, No, Yes, Yes, Yes, Yes, No, No, No, Yes, No, Yes.

A. Point Estimate (p)

  • Formula: p = \frac{x}{n} where:
    • x = number of "Yes" responses
    • n = total number of responses
  • Calculation:
    • p = \frac{23}{40} = 0.575

B. Margin of Error (MOE)

  • Margin of Error: generally determined by MOE = z_{\alpha/2} \times \sqrt{\frac{p(1-p)}{n}}
  • Using a standard normal distribution for a 95% confidence level:
    • z_{\alpha/2} = | \text{Norm}(0.05, 0, 1) | = 1.96
  • Calculation:
    • For p:
    • MOE = 1.96 \times \sqrt{\frac{0.575(1-0.575)}{40}}
    • After evaluation, MOE = 0.1532

C. Sample Size

  • Condition: Is the sample size large? Need to check:
    • Condition: np \geq 10 and n(1-p) \geq 10.
  • For calculation:
    • n \cdot p = 40 \cdot 0.575 = 23 \geq 10
    • n(1-p) = 40 \cdot 0.425 = 17 \geq 10
  • Concluded: Yes, the sample size is large.

D. 95% Confidence Interval

  • Formula involves using the point estimate and margin of error:
  • CI = (p - MOE, p + MOE)
  • Result Calculation:
    • CI = (0.575 - 0.1532, 0.575 + 0.1532) = (0.4218, 0.7282)

E. Interpretation

  • Conclusion: "I am 95% confident that the true proportion (p) of adults who drink occasionally is between 42.18% and 72.82%."

Example 6.13 - Confidence Interval for Mean Pulse Rate

  • Context: A random sample of 21 US adult males who jog at least 15 miles per week; average pulse rate measured.
  • Sample Average Pulse Rate: 52.6 beats/minute.
  • Standard Deviation: 3.22 beats/minute.
  • Distribution: Assumed to be normally distributed.

A. 95% Confidence Interval Calculation

  • Sample Size: n = 21
  • Critical Value (z): $z_{0.025} = 2.04$ obtained from t-distribution table.
  • Margin of Error (ME):
    • ME = z_{0.025} \cdot \frac{s}{\sqrt{n}}
  • Full Calculation:
    • ME = 2.04 \cdot \frac{3.22}{\sqrt{21}} = 1.469
  • Resulting Interval:
    • CI = (52.6 - 1.469, 52.6 + 1.469) = (51.13, 54.07)

C. Interpretation of Results

  • Conclusion: "We are 95% confident that the true mean pulse rate of US adult males who jog at least 15 miles per week is between 51.13 and 54.07."

D. Reduce Observation Interpretation

  • Comparison: Mean pulse rate of all US adult males at approximately 72 beats/minute is outside the confidence interval, indicating jogging likely reduces mean pulse rate.

Example 7.10 - Hypothesis Testing for Visits to Physician

  • Context: Gallup Poll reports average visits to physician by women.
  • Null Hypothesis: H_0: \mu = 5.8 visits/year.
  • Sample Size: 12 women showed visits data: 3, 2, 1, 3, 7, 2, 9, 4, 6, 6, 8, 5.

A. Statistical Calculation by Hand

  • Test Statistic (t):
    • t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
  • For this sample: Mean = \bar{x} = 4.667, Standard Deviation (s) calculated.
  • Result:
    • t = -1.527485

B. Use of R to Calculate t-test

  • Command in R:
    • t.test(visits, alternative='two.sided', mu=5.8, conf.level=0.95)
  • Result Analysis:
    • t = -1.5275, df = 11, p-value = 0.1549
  • Interpretation indicated alternative hypothesis tested: True mean is not equal to 5.8.
  • 95% confidence interval calculated: (3.033623, 6.299710).

Example 7.11 - Hypothesis Test on Field Trip Interest

  • Context: A teacher hypothesizes that 85% of students want to visit the zoo.
  • Sample Size: 55 students, 39 expressed interest.

Formulating Hypothesis

  • Null Hypothesis (H0):
    • H_0: p = 0.85
  • Alternative Hypothesis (H1):
    • H_1: p
      eq 0.85 (two-tailed)

Example 7.12 - Writing Hypotheses

Practice Writing the Null and Alternative Hypotheses

A. Pregnancies in the country not planned:

  • Null in words: Half of the pregnancies in this country are not planned.
  • Null in symbols: H_0: p = 0.5
  • Direction: Left-tailed.
  • Alternative in symbols: H_1: p > 0.5

B. Chicken pox vaccination effectiveness:

  • Null in words: 3% of vaccinated children still contract the disease.
  • Null in symbols: H_0: p = 0.03
  • Direction: Left-tailed.
  • Alternative in symbols: H_1: p < 0.03

C. Lactose intolerance in population:

  • Null in words: Proportion of lactose-intolerant people is 0.25.
  • Null in symbols: H_0: p = 0.25
  • Direction: Two-tailed.
  • Alternative in symbols: H_1: p
    eq 0.25

Example 7.17 - Polling Majority Preferences

  • Context: Poll of 1000 adult Americans on political preference, 48% prefer a Democratic candidate.
  • Null Hypothesis:
    • H_0: p = 0.5
  • Sample Size Check: Confirmed: large sample.
    • Test statistic and p-value calculated:
    • z = -1.26, p-value = 0.2059
  • Conclusion: Fail to reject H_0; insufficient evidence to state that true proportion differs from 50%.

Example 8.3 - Weight Change in Smokers

  • Context: Investigation of weight change after quitting smoking among eight females.
  • Results Pre and Post Measurements:
    • Before/After differences calculated for weight.

A. Point Estimate Calculation

  • Point Estimate for Difference: P = 1.75

B. Margin of Error Calculation

  • Margin of Error: M.E. = 2.853

C. Confidence Interval Calculation

  • Resulting interval is (-1.103, 4.6026)

D. Interpretation of Confidence Interval

  • Conclusion: "We are 95% confident that the true mean difference in women's weight before and after quitting smoking is between -1.103 and 4.6026."

Example 8.3 Continued - Paired T-Test

  • Significance Level for Test: 5%.
  • Test Hypotheses:
    • H0: \mud = 0
    • Two-tailed test conducted.
  • Result:
    • Decision: Fail to reject the null hypothesis; conclusions indicated insufficient evidence to suggest significant weight change post quitting smoking.