Question

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???, abbreviated as SEM, rather than the standard deviation s.) Group n x¯¯¯ SEM Alcohol 29 0.23 0.05 Placebo 29 0.1 0.03 What are the standard deviations (±0.001) for the two groups? For "Alcohol" group s1= For "Placebo" group s2= What degrees of freedom does the conservative Option 2 use for two-sample t procedures for these samples? Using Option 2, a 95% confidence interval (±0.01) for the mean difference (Alcohol - Placebo) between the two groups is from to

Answer 1

Solution:

From given information, we have

Group	n	Xbar	SEM
Alcohol	29	0.23	0.05
Placebo	29	0.1	0.03

SE or SEM = S/sqrt(n)

For alcohol, SEM = 0.05, n1 = 29

So, 0.05 = S/sqrt(29)

S = 0.05*sqrt(29) = 0.26925824

Standard deviation for Alcohol group = S1 = 0.269

For placebo group, SEM = 0.03, n2 = 29

So, 0.03 = S/sqrt(29)

S = 0.03*sqrt(29) = 0.161554944

Standard deviation for Placebo group = S2 = 0.162

So, we have

Group	n	Xbar	S
Alcohol	29	0.23	0.269
Placebo	29	0.1	0.162

Now, we have to conduct two sample t test.

H₀: µ₁ = µ₂

vs.

H_a: µ₁ ? µ₂

(Two tailed test)

Total degrees of freedom = n1 + n2 – 2 = 29 + 29 – 2 = 58 – 2 = 56

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(29 – 1)*0.269^2 + (29 – 1)*0.162^2]/(29 + 29 – 2)

S_p² = 0.0493

t = (0.23 – 0.10) / sqrt[0.0493*((1/29)+(1/29))]

t = 0.13 / sqrt[0.0493*((1/29)+(1/29))]

t = 0.13/0.0583

t = 2.2294

P-value = 0.0298

(by using t-table or excel)

? = 0.05

P-value < ? = 0.05

So, we reject the null hypothesis

Now, we have to find 95% confidence interval for difference between two group means.

Confidence level = 95%

df = 56

Critical t-value = 2.0032

Confidence interval = (X1bar – X2bar) -/+ t* sqrt[S_p²*((1/n1)+(1/n2))]

Confidence interval = 0.13 -/+ 2.0032* 0.0583

Confidence interval = 0.13 -/+ 0.1168

Lower limit = 0.13 – 0.1168 = 0.0132

Upper limit = 0.13 + 0.1168 = 0.2468

Confidence interval = (0.0132, 0.2468)

A 95% confidence interval (±0.01) for the mean difference (Alcohol - Placebo) between the two groups is from 0.01 to 0.25.