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 23.6 with a standard deviation of 4.5​, while the 200 students in group 2 had a mean score of 17 with a standard deviation of 3.9. Complete parts ​(a) and ​(b) below.

a.) Determine the 90​% confidence interval for the difference in​ scores, μ1−μ2. Interpret the interval.

the upper bound is ?

the lower bound is ?

Answer 1

i am using minitab to solve the problem.

steps:-

stat basic statistics two sample t   select summarized data from the drop down menuin sample 1 type 200 in sample size , 23.6 in sample maen and 4.5 in sample sd in sample 2 type 200 in sample size , 17 in sample maen and 3.9 in sample sd options in confidence level type 90in hypothesized difference type 0select alternative hypothesis as difference hypothesized difference.okok.

your needed output be:-

Estimation for Difference

Difference	90% CI for Difference
6.600	(5.906, 7.294)

**SOLUTION**

the 90​% confidence interval for the difference in​ scores, μ1−μ2 is:-

upper bound = 7.294

lower bound = 5.906

interpration:-

we are 90% confident that the true mean difference between group 1 and group 2 will lie within 5.906 and 7.294 .

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