Question

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a​ professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 22.8 with a standard deviation of 3.5​, while the 200 students in group 2 had a mean score of 15.3 with a standard deviation of 2.6.

Complete parts ​(a) and ​(b) below.

​(a) Determine the

95​% confidence interval for the difference in​ scores,

mu 1 minus mu 2μ1−μ2.

Interpret the interval.

Lower bound is:

Upper bound is:

B)What does this say about​ priming?

Answer 1

since sample size is large, we can use normal distribution (z) for above,

x1            =	22.80	x2            =	15.30
n1           =	200	n2           =	200
σ1           =	3.50	σ2           =	2.60
std error σx1-x2=√(σ21/n1+σ22/n2)    =			0.308

Point estimate of differnce '=x1-x2 =			7.500
for 95 % CI value of z=			1.960
margin of error E=z*std error =			0.604
lower bound=(x1-x2)-E =			6.90
Upper bound=(x1-x2)+E =			8.10

b)

since interval values are above 0, this tells us that group 1 has higher mean then group 2.