BIOSTATS 3.1 - CH. 17: INFERENCE ABOUT A POPULATION MEAN

• Study notes for Biostatistics & Statistical Analysis, Chapter 17: Inference for a population mean (σ unknown)

Inference Conditions

• Data must come from a simple random sample.
• The population observations must be normally distributed.
• The sample mean x̄ has a normal distribution with a standard deviation of σ/√n.
• When σ is unknown, it is estimated using s; the estimated standard deviation of x̄, known as the standard error, is s/√n.

T Distributions

• Considers a random sample of size n from a Normal population N(µ,σ).
• When s is known, the sampling distribution of x̄ is Normal N(µ, σ/√n), and the statistic z = (x̄-µ) / (σ/√n) follows the standard Normal N(0, 1).
• When σ is estimated from the sample standard deviation s, the statistic t = (x̄-µ) / (s/√n) follows the t distribution with n - 1 degrees of freedom.
• When n is small, the t distribution df n-1 has more area in the tails than the Standard Normal distribution.
• When n is large, s is a good estimate of σ, and the t distribution df n-1 is close to the standard Normal distribution.

Standard Deviation vs. Standard Error

• The sample standard deviation s for a sample size n has n-1 degrees of freedom.
• The formula for sample standard deviation s is s = √[Σ(xi - x̄)² / (n-1)].
• The value s/√n is called the standard error of the mean (SEM).
• In a medical study, a new medication's effect on seated systolic blood pressure for 25 patients is presented as mean ± SEM, which is 113.5 ± 8.9, the standard deviation s of the sample data is calculated as s = SEM * √n = 8.9 * √25 = 44.5.
• Table C displays z-values and t-values corresponding to landmark P-values and confidence levels.

One-Sample t Confidence Interval

• A confidence interval is a range of values that contains the true population parameter with probability C (confidence level).
• For a population with unknown µ and σ, x̄ estimates µ, and s estimates σ, using a t distribution (df n - 1).
• C represents the area between -t* and t*.
• To find t*, refer to Table C.
• The margin of error m is calculated as m = t* * (s/√n).
• Data on blood cholesterol levels (mg/dL) of 24 lab rats show a sample mean of 85 and a standard deviation of 12; a 95% confidence interval for the mean blood cholesterol of all lab rats is calculated by first finding df = n - 1 = 24 - 1 = 23.
• Then calculating m = t* * (s/√n) = 2.069 * (12/√24) = 5.07
• This provides x̄ ± m = 85 ± 5.1, resulting in a range of 79.9 to 90.1 mg/dL.
• We are 95% confident that the true mean blood cholesterol of all lab rats is between 79.9 and 90.1 mg/dL.

One-Sample t-Test

• A test of hypotheses involves stating the null hypothesis (Ho), deciding on a one-sided or two-sided alternative (Ha), and choosing a significance level α.
• Steps include calculating t and its degrees of freedom, finding the area under the curve with Table C or software, and stating the P-value and conclusion.
• We draw a random sample of size n from an N(µ, σ) population.
• When s is estimated from the sample, the distribution of the test statistic t is a t distribution with df = n - 1.
• The formula is t = (x̄ - µ0) / (s/√n), and H0: µ = µ0.
• Resulting t-test is robust to deviations from Normality when the sample size is sufficiently large.
• The P-value represents the probability, if H0 were true, of randomly drawing a sample like the one obtained or more extreme in the direction of Ha.
• Using Table C, for Ha: µ > µ0, if n = 10 and t = 2.70, then 2.398 < t = 2.7 < 2.821, so 0.02 > P-value > 0.01.

T-Test Example

• Participants in a study were 53 obese children aged 9-12, with a BMI above the 95th percentile for their age and gender.
• The intervention involved family counseling sessions on the stoplight diet (green/yellow/red approach to eating food) over 8 weekly sessions and 3 follow-up sessions, and weight change was assessed at 15 weeks of intervention.
• The study aimed to determine if the intervention effectively helped obese children lose weight, where Ho: µ = 0 versus Ha: µ < 0 (one-sided test), and N=53, Mean = -2.404, SE Mean = 0.720, and StDev = 5.243.
• For df = 52 ≈ 50, 3.261 < |t| = 3.34 < 3.496, indicating 0.001 > one-sided P > 0.0005.
• The software gives P = 0.0008 ≈ 0.001, which is highly significant, indicating a significant weight loss on average, following the intervention.

Matched Pairs T-Procedures

• Compares treatments or conditions at the individual level, where data sets produced are not independent and individuals in one sample are related to those in the other sample.
• Pre-test and post-test studies look at data collected on the same sample elements before and after some experiment is performed.
• Twin studies often try to sort out the influence of genetic factors by comparing a variable between sets of twins.
• Using people matched for age, sex, and education in social studies allows canceling out the effect of these potential lurking variables.
• Paired data is used to test for the difference in the two population means.
• The variable studied becomes x̄diff, average difference, and H0: µdiff = 0; Ha: µdiff > 0 (or < 0, or ≠ 0).
• Conceptually, this is just like a test for one population mean.
• Participants in the example study were 53 obese children ages 9 to 12 with a BMI above the 95th percentile for age and gender.
• The intervention involved family counseling sessions on the stoplight diet (green/yellow/red approach to eating food) after 8 weekly sessions and 3 follow-up sessions.
• Weight change was assessed at 15 weeks of intervention to see if the intervention was effective in helping obese children lose weight.
• In the weight change values, the difference in body weight is before and after intervention for each participant.

Caffeine Deprivation

• Explores if lack of caffeine increases depression.
• In a study, randomly selected caffeine-dependent individuals were deprived of all caffeine-rich foods and assigned to receive daily pills, those pills at one time contained caffeine and, at another time, a placebo; depression was assessed quantitatively.
• The experiment used a matched pairs design with 2 measurements for each subject.
• "Difference" is computed by Placebo minus Caffeine.
• With 11 "difference" points, df = n − 1 = 10; Xdiff = 7.36; Sdiff = 6.92; SEMdiff = Sdiff / √n = 6.92/√11 = 2.086.
• Results are tested to determine whether H0: mdiff = 0; Ha: mdiff > 0.
• t = (x̄diff - µdiff) / (Sdiff/√n) = (x̄diff - 0) / SEMdiff = 7.36 / 2.086 ≈ 3.53.
• For df = 10, 3.169 < t < 3.53 < 3.581 indicates 0.005 > P-value > 0.0025 (Software gives P = 0.0207).
• Caffeine deprivation causes a significant increase in depression (P < 0.005, n = 11).

Robustness

• T procedures are exactly correct when the population is exactly Normal, but this is rare.
• T procedures are robust to small deviations from Normality, but the sample must be a random sample from the population.
• Outliers and skewness strongly influence the mean and therefore the t procedures.
• The impact diminishes as the sample size gets larger because of the central limit theorem.
• As a guideline:
  • When n < 15, the data must be close to Normal and without outliers.
  • When 15 > n > 40, mild skewness is acceptable, but not outliers.
  • When n > 40, the t statistic will be valid even with strong skewness.

T-Procedures: Calcium Absorption

• Explores if oligofructose consumption stimulates calcium absorption.
• Healthy adolescent males took a pill for nine days and had their calcium absorption tested on the ninth day, experiment was repeated three weeks later, subjects received either an oligofructose pill first or a control sucrose pill first and the order was randomized and the experiment was double-blind.
• The study gathered fractional calcium absorption data (in percent of intake) for 11 subjects.

T-Procedures: Red Wine

• Examines if drinking red wine increases blood polyphenol levels, thus maybe protecting against heart attacks.
• Nine randomly selected healthy men were assigned to drink half a bottle of red wine daily for two weeks.
• Assessed: the percent change in their blood polyphenol levels.
• x̄= 5.5; s = 2.517; df = n - 1 = 8