Question

We considered the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey in Exercise 5.21. The mean and standard deviation of the differences are x̄_read-write = -0.545 and 8.887 points respectively.

(a) Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students.
lower bound: points (please round to two decimal places)
upper bound: points (please round to two decimal places)

Interpret this interval in context.

• We can be 95% confident that our confidence interval contains the mean difference between reading and writing scores of these 200 students
• 95% of students will have a difference between reading and writing scores that falls within our confidence interval
• We can be 95% confident that the average difference between reading and writing scores of all students is contained within our confidence interval

(c) Does the confidence interval provide convincing evidence that there is a real difference in the average scores? Explain.

• no, because or confidence interval contains both positive and negative values
• yes, because negative scores are impossible and our confidence interval contains them
• no, since 0 is contained in our confidence interval
• yes, since 0 is contained in our confidence interval

Answer 1

Solution :

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z _0.025 = 1.96

Lower bound = -0.545 - 1.96 * 8.887 = -17.96

Upper bound = -0.545 + 1.96 * 8.887 = 16.87

We can be 95% confident that the average difference between reading and writing scores of all students is contained within

our confidence interval

no, because or confidence interval contains both positive and negative values