Question

A counselor at the local chapter of Impulse Buyers Anonymous had just conducted a weekend workshop on how to be less suggestible. To find out if the workshop had any effect, she had each participant fill out a questionnaire that measured suggestibility, with high scores indicating high suggestibility. (Several such questionnaires exist.) Suppose that the national norm (mean) for the questionnaire that was used was 25 and that the participants produced the following statistics. Construct a 95 percent confidence interval and sentence of interpretation that tells about the effect of the workshop.ΣΧ = 242 ΣΧ2 = 5,413 N = 11

Answer 1

(20.322 ; 23.678)

Explanation:

Given that :

ΣΧ = 242 ; ΣΧ2 = 5,413 ; N = 11

Mean, m = ΣΧ / N = 242 / 11 = 22

Standard deviation = sqrt[(ΣΧ2/N) - (ΣΧ/N)²]

Standard deviation = sqrt[(5413/11) - (242/11)^2]

Standard deviation = sqrt(8.0909090) = 2.84

Confidence interval = m ± Zcritical * s/sqrt(n)

Zcritical at 95% = 1.96

Lower boundary = 22 - 1.96*2.84/sqrt(11)

Lower boundary = 22 - (1.96 * 0.8562922) = 20.322

Upper boundary = 22 + (1.96 * 0.8562922) = 23.678