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 200students in group 1 had a mean score of 24.5 with a standard deviation of 4.3​, while the 200 students in group 2 had a mean score of 16.3 with a standard deviation of 3.1.Complete parts ​(a) and ​(b) below.​(a) Determine the 90% confidence interval for the difference in​ scores, mu 1 minus mu 2.Interpret the interval. ​( , ) ​(Round to three decimal places as​needed.)Interpret the interval. Choose the correct answer below.A.There is a 90% probability that the difference of the means is in the interval.B.The researchers are 90% confident that the difference of the means is in the interval.C.The researchers are 90% confident that the difference between randomly selected individuals will be in the interval.D.There is a 90​% probability that the difference between randomly selected individuals will be in the interval.​(b) What does this say about​ priming?A.Since the 90% confidence interval does not contain​ zero, the results suggest that priming does not have an effect on scores.B.Since the 90% confidence interval does not contain​ zero, the results suggest that priming does have an effect on scores.C.Since the 90% confidence interval contains​ zero, the results suggest that priming does have an effect on scores.D.Since the 90% confidence interval contains​ zero, the results suggest that priming does not have an effect on scores.

Answer 1

Explanation:

a) The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean of group 1

x2 = sample mean of group 2

s1 = sample standard deviation for data 1

s2 = sample standard deviation for data 2

For a 90% confidence interval, the z score is 1.645

From the information given,

x1 = 24.5

s1 = 4.3

n1 = 200

x2 = 16.3

s2 = 3.1

n2 = 200

x1 - x2 = 24.5 - 16.3 = 8.2

z√(s1²/n1 + s2²/n2) = 1.645√(4.3²/200 + 3.1²/200) = 1.645√0.1405

z = 0.62

Therefore, the 90% confidence interval is 8.2 ± 0.62

Interpretation:

B.The researchers are 90% confident that the difference of the means is in the interval.

b) B.Since the 90% confidence interval does not contain​ zero, the results suggest that priming does have an effect on scores.