Question

1. Senior finance majors were randomly assigned to a virtual team that met only electronically, or to a face-to-face team that met in person.  Both teams were presented with a task of 8 cases, and afterwards they completed a survey.  On the 1-5 Likert scale prompt “As compared to other teams, the members got along together”, team v had a mean score of 2.48 with a standard deviation of .76 and n=44, while team f had a mean score of 1.83 with a standard deviation of .82 and n=42.  Find a 90% confidence interval for the mean difference. What does the CI indicate about the mean difference? Please show all work.

Answer 1

Sol:

90% confidence interval for difference in meansis

(x1bar-x2bar)+-tcrit*sqrt(s1^2/n1+s2^2/n2)

t crit for n1+n2-2 df

df=n1+n2-2=44+42-2=84

alpha=0.10

alpha/2=0.10/2=0.05

t critica l in Excel

=T.INV(0.05,84)

=1.6632

(2.48-1.83)+-1.6632*sqrt((0.76^2/44)+(0.82^2/42))

0.65- 0.2839,0.65+0.2839

0.3661,0.9339

0.3661<mu1-mu2<0.9339

Ho:mu1-mu2=0

Ha:mu1-mu2 not =0

Since this 90% confidence interval for mean do not contain zero

Reject Ho

accept Ha

There is sufficient statistical by 90% confidence interval that there are difference in mean scores of team v and team f