10.

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 24.2 with a standard deviation of 3.9, while the 200 students in group 2 had a mean score of 16.3 with a standard deviation of 3.7. Complete parts (a) and (b) below.

(a)Determine the 90% confidence interval for the difference in scores, 1 2 . Interpret the interval.

(,)

(Round to three decimal places as needed.)

Interpret the interval. Choose the correct answer below.

A.

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

B.There is a 90% probability 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 contains zero, the results suggest that priming does have an effect on scores.

B.

Since the 90% confidence interval contains zero, the results suggest that priming does not have an effect on scores.

C.

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

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

11.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.91 hours, with a standard deviation of 2.44 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.24 hours, with a standard deviation of 1.87 hours. Construct and interpret a 90% confidence interval

for the mean difference in leisure time between adults with no children and adults with children

1 2.

Let 1 represent the mean leisure hours of adults with no children under the age of 18 and 2 represent the mean leisure hours of adults with children under the age of 18.

The 90% confidence interval for

1 2

is the range from hours tohours.

(Round to two decimal places as needed.)

What is the interpretation of this confidence interval?

A.

There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours.

B.

There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

C.

There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

D.There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours.