The Practice of Statistics in the Life Sciences Fourth Edition Chapter 17: Inference for a population mean (σ unknown) Copyright © 2018 W. H. Freeman and Company

Objectives Inference for the mean of one population (σ unknown) Conditions for inference The t distributions The one-sample t confidence interval The one-sample t test Matched pairs t procedures Robustness

Conditions for inference The data must come from a simple random sample. The observations in the population must have a Normal distribution. The sample mean 𝑥 has a Normal distribution with standard deviation σ /√n. When σ is unknown, we estimate it using s. The estimated standard deviation of 𝑥 , known as the standard error, is then s/√n.

The t distributions We take a random sample of size n from a Normal population N(µ,σ): When s is known, the sampling distribution of 𝑥 is Normal N 𝜇, 𝜎 𝑛 , and the statistic 𝑧= 𝑥 −𝜇 𝜎 𝑛 follows the standard Normal 𝑁 0, 1 . When 𝜎 is estimated from the sample standard deviation s, the statistic 𝑡= 𝑥 −𝜇 𝑠 𝑛 follows the t distribution with 𝑛−1 degrees of freedom.

The t distribution vs the Normal (1 of 2) t distributions have more area in the tails than the standard Normal distribution. t with infinite df is in fact N(0,1). When n is small, t df n – 1 distribution has more area in the tails than does the Standard Normal distribution.

The t distribution vs the Normal (2 of 2) t distributions have more area in the tails than the standard Normal distribution. t with infinite df is in fact N(0,1). When n is large, s is a good estimate of s and the t df n – 1 distribution is close to the standard Normal distribution.

Standard deviation vs standard error (1 of 2) For a sample of size n, the sample standard deviation s is: n − 1 is the “degrees of freedom.” 𝑠= 1 𝑛−1 𝑥 𝑖 − 𝑥 2 The value s/√n is called the standard error of the mean SEM. Scientists often present their sample results as the mean ± SEM.

Standard deviation vs standard error (2 of 2) A medical study examined the effect of a new medication on the seated systolic blood pressure. The results, presented as mean ± SEM for 25 patients, are 113.5 ± 8.9. What is the standard deviation s of the sample data? SEM = s/√n <=> s = SEM*√n s = 8.9*√25 = 44.5

Table C

The one-sample t confidence interval (1 of 2) A confidence interval is a range of values that contains the true population parameter with probability (confidence level) C. We have a set of data from a population with both  and α unknown. We use 𝑥 to estimate , and s to estimate σ, using a t distribution (df n − 1). C is the area between −t* and t*. We find t* in the line of Table C. The margin of error m is: 𝑚= 𝑡 ∗ 𝑠 𝑛 When using software or a calculator, you won’t need to obtain the value of t*; this will be done automatically by the program.

The one-sample t confidence interval (2 of 2) When using software or a calculator, you won’t need to obtain the value of t*; this will be done automatically by the program.

Confidence interval example (1 of 2) Data on the blood cholesterol levels (mg/dL) of 24 lab rats give a sample mean of 85 and a standard deviation of 12. We want a 95% confidence interval for the mean blood cholesterol of all lab rats. 𝑑𝑓=𝑛−1=24−1=23 𝑚= 𝑡 ∗ 𝑠 𝑛 =2.069 12 24 =5.07 𝑥±𝑚=85±5.1, or 79.9 to 90.1 mg/dL

Confidence interval example (2 of 2) We are 95% confident that the true mean blood cholesterol of all lab rats is between 79.9 and 90.1 mg/dL.

The one-sample t test (1 of 3) As before, a test of hypotheses requires a few steps: Stating the null hypothesis (H0) Deciding on a one-sided or two-sided alternative (Ha) Choosing a significance level α Calculating t and its degrees of freedom Finding the area under the curve with Table C or software Stating the P-value and the conclusion

The one-sample t test (2 of 3) We draw a random sample of size n from an N(µ, σ) population. When s is estimated from s, the distribution of the test statistic t is a t distribution with df = n – 1. 𝐻 0 : 𝜇=𝜇 𝑡= 𝑥 − 𝜇 0 𝑠 𝑛 This resulting t test is robust to deviations from Normality as long as the sample size is large enough.

The one-sample t test (3 of 3) The P-value is the probability, if H0 was true, of randomly drawing a sample like the one obtained or more extreme in the direction of Ha. The P-value corresponds to the one-tailed or two-tailed (depending on Ha) area under the t (df n – 1) curve.

Using Table C: For Ha: μ > μ0 if n = 10 and t = 2.70, then…

t test example (1 of 4) Study Participants: 53 obese children ages 9 to 12 with a BMI above the 95th percentile for age and gender Intervention: family counseling sessions on the stoplight diet (green/yellow/red approach to eating food)—after 8 weekly sessions and 3 follow-up sessions Assessment: weight change at 15 weeks of intervention Was the intervention effective in helping obese children lose weight? H0:  = 0 versus Ha:  < 0 (one-sided test) Variable N Mean SE Mean StDev Weightchange 53 -2.404 0.720 5.243

t test example (2 of 4)

t test example (3 of 4) MINITAB: Test of mu = 0 vs < 0 Variable N Mean StDev SE Mean T P Weightchange 53 -2.404 5.243 0.720 -3.34 0.001

t test example (4 of 4) For df = 52 ≈ 50, 3.261 < |t| = 3.34 < 3.496,  0.001 > one-sided P > 0.0005 (software gives P = 0.0008 ≈ 0.001), highly significant. There is a significant weight loss, on average, following intervention.

Matched pairs t procedures (1 of 2) Sometimes we want to compare treatments or conditions at the individual level. The data sets produced this way are not independent. The 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.

Matched pairs t procedures (2 of 2) Using people matched for age, sex, and education in social studies allows us to cancel out the effect of these potential lurking variables. In these cases, we use the paired data to test for the difference in the two population means. The variable studied becomes 𝑥 𝑑𝑖𝑓𝑓 , average difference, and H0: µdiff = 0; Ha: µdiff > 0 (or < 0, or ≠ 0) Conceptually, this is just like a test for one population mean.

Matched pairs example (1 of 2) Study Participants: 53 obese children ages 9 to 12 with a BMI above the 95th percentile for age and gender Intervention: family counseling sessions on the stoplight diet (green/yellow/red approach to eating food)—after 8 weekly sessions and 3 follow-up sessions Assessment: Weight change at 15 weeks of intervention Was the intervention effective in helping obese children lose weight? This is a pre-/post design. The weight change values are the difference in body weight before and after intervention for each participant.

Matched pairs example (2 of 2) Variable N Mean SE Mean StDev Weightchange 53 -2.404 0.720 5.243

Another matched pairs example (1 of 3) Does lack of caffeine increase depression? Randomly selected caffeine-dependent individuals were deprived of all caffeine-rich foods and assigned to receive daily pills. At one time the pills contained caffeine and, at another time they were a placebo. Depression was assessed quantitatively (higher scores represent greater depression).

Another matched pairs example (2 of 3) On the following slide is a matched pairs design with 2 measurements for each subject. We compute a new variable “Difference” Placebo minus Caffeine

Another matched pairs example (3 of 3)

Another matched pairs example (1 of 2) With 11 "difference" points, df = n – 1 = 10. We find: 𝑥 𝑑𝑖𝑓𝑓 = 7.36; sdiff = 6.92; so SEMdiff = sdiff / √n = 6.92/√11 = 2.086 We test: H0: mdiff = 0 ; Ha: mdiff > 0 𝑡= 𝑥 𝑑𝑖𝑓𝑓 − 𝜇 𝑑𝑖𝑓𝑓 𝑠 𝑑𝑖𝑓𝑓 𝑛 = 𝑥 𝑑𝑖𝑓𝑓 −0 𝑆𝐸𝑀 𝑑𝑖𝑓𝑓 = 7.36 2.086 ≈3.53

Another matched pairs example (2 of 2) For df = 10, 3.169 < t 3.53 < 3.581  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 (1 of 2) The t procedures are exactly correct when the population is exactly Normal. This is rare. The 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. Their impact diminishes as the sample size gets larger because of the central limit theorem.

Robustness (2 of 2) 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.

Are t procedures appropriate? (1 of 4) Does oligofructose consumption stimulate calcium absorption? Healthy adolescent males took a pill for nine days and had their calcium absorption tested on the ninth day. The experiment was repeated three weeks later. Subjects received either an oligofructose pill first or a control sucrose pill first. The order was randomized and the experiment was double-blind. Fractional calcium absorption data (in percent of intake) for 11 subjects: The data are from a randomized comparative experiment. The Normal quantile plot suggest Normally distributed data. So a t test would be appropriate here.

Are t procedures appropriate ? (2 of 4) The data are from a randomized comparative experiment. The Normal quantile plot suggest Normally distributed data. So a t test would be appropriate here. Can we use a t inference procedure for this study? Discuss the assumptions.

Are t procedures appropriate ? (3 of 4) Red wine, in moderation Does drinking red wine in moderation increase 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. The percent change in their blood polyphenol levels was assessed: The data are from a randomized comparative experiment. However, the dotplot shows a low outlier in this small data set. This suggests that the population might not be Normally distributed. Given the small sample size, a t test may not be appropriate here.

Are t procedures appropriate (4 of 4) 0.7 3.5 4 4.9 5.5 7 7.4 8.1 8.4 𝑥 = 5.5; s = 2.517; df = n − 1 = 8 The data are from a randomized comparative experiment. However, the dotplot shows a low outlier in this small data set. This suggests that the population might not be Normally distributed. Given the small sample size, a t test may not be appropriate here. Can we use a t inference procedure for this study? Discuss the assumptions.