Question

We consider the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey. The mean and standard deviation of the differences are

xread − write = −0.537
and 8.882 points.
(a)
Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students. (Round your answers to two decimal places.)

(b)
Interpret this interval in context. (Round your answers to two decimal places.)
We are 95% confident that on the reading test students score, on average,
points lower to points higher than they do on the writing test.

Answer 1

(a) To calculate a 95% confidence interval for the average difference, we use the formula: mean ± (1.96 * (standard deviation / sqrt(sample size))). Here, 1.96 is the z-score for a 95% confidence interval.

So, the calculation would be: -0.537 ± (1.96 * (8.882 / sqrt(200)))

The confidence interval would be approximately -1.22 to 0.15 (rounded to two decimal places).

(b) We are 95% confident that on the reading test students score, on average, 1.22 points lower to 0.15 points higher than they do on the writing test.