Question

(21.28) Healthy men aged 21 to 35 were randomly assigned to one of two groups: half received 0.82 grams of alcohol per kilogram of body weight; half received a placebo. Participants were then given 30 minutes to read up to 34 pages of Tolstoy's War and Peace (beginning at chapter 1, with each page containing approximately 22 lines of text). Every two to four minutes participants were prompted to indicate whether they were "zoning out." The proportion of times participants indicated they were zoning out was recorded for each subject. The table below summarizes data on the proportion of episodes of zoning out.

(The study report gave the standard error of the mean s/n−−√s/n, abbreviated as SEM, rather than the standard deviation s.)

Group n x¯¯¯x¯ SEM
Alcohol 22 0.23 0.06
Placebo 22 0.14 0.02

What are the standard deviations (±±0.001) for the two groups?

For "Alcohol" group s1=s1=

For "Placebo" group s2=s2=

What degrees of freedom does the conservative Option 2 use for two-sample t procedures for these samples?

Using Option 2, a 90% confidence interval (±±0.01) for the mean difference (Alcohol - Placebo) between the two groups is  from to .

Answer 1