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 26.3 with a standard deviation of 4​, while the 200 students in group 2 had a mean score of 18.7 with a standard deviation of 3.3. Complete parts ​(a) and ​(b) below

(a) Determine the 95​%confidence interval for the difference in​ scores, μ1−μ2. Interpret the interval.

​(b) What does this say about​ priming?

Answer 1

solution:-

given that

group 1

x1 = 26.3 , s1 = 4 , n1 = 200

group 2

x2 = 18.7 , s2 = 3.3 , n2 = 200

degree of freedom df = (n1+n2)-2 = (200+200)-2 = 398

we look into t table with df = 398 and with 95% confidence level

critical value t = 1.966

(a) confidence interval

formula

=> (x1-x2) +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> (26.3-18.7) +/- 1.966 * sqrt((4^2/200) + (3.3^2/200))

=> (6.879 , 8.321)

Interpret the interval

=> the researchers are 95% confident that the difference of the means is in the interval

(b) What does this say about​ priming

=> since the 95% confidence interval does not contain zero, thr results suggest that priming does have an effect on scores