Question

High School and Beyond, Part II: 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 5.3. The mean and standard deviation of the differences are is = -0.545 and 8,887 points.

a) Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students.

b) Interpret this interval in context.

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

Answer 1

High School and Beyond, Part II: 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 5.3. The mean and standard deviation of the differences are is \(=-0.545\) and 8,887 points.

1. Calculate a \(95 \%\) confidence interval for the average difference between the reading and writing scores of all students.

For the sample size of \(200, z\) distribution used

$$ \begin{aligned} &\mathrm{N}=200 \\ &\text { Mean }=-0.545 \\ &\text { Sd }=8.887 \\ &\quad \bar{x} \pm z * \frac{s}{\sqrt{n}} \\ &-0.545 \pm 1.96 * \frac{8.887}{\sqrt{200}} \\ &=(-1.777,0.687) \end{aligned} $$

1. Interpret this interval in context.

We are \(95 \%\) confident that average difference between the reading and writing scores of all students falls in the interval \((-1.777,0.687)\)

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

The \(95 \%\) confidence interval contains zero value. There fore null hypothesis is not rejected.

The confidence interval provides that there is no evidence that there is a real difference in the average scores.