Question

A group of 56 computer science students were taught introductory computer programming class with an innovative teaching method that used a graphical interface and drag-and-drop methods of creating computer programs. At the end of the class, 43 of these students said that they felt confident in their ability to write computer programs. Another group of 41 students were taught the same material using a standard method. At the end of class, 25 of these students said they felt confident. Assume that each class contained a simple random sample of students. Let pX represent the population proportion of students taught by the innovative method who felt confident and let pY represent the population proportion of students taught by the standard method who felt confident. Find a 99% confidence interval for the difference pX−pY . Round the answers to four decimal places.

Answer 1

(-0.0861, 0.4023)

Explanation:

We have large sample sizes
`n_(x) = 56` and
`n_(y) = 41`. A
`100(1-\alpha)`% confidence interval for the difference
`p_(x)-p_(y)` is given by
`(\hat{p}_(x)-\hat{p}_(y))\pm z_(\alpha/2)\sqrt{\frac{\hat{p}_(x)(1-\hat{p}_(x))}{n_(x)}+\frac{\hat{p}_(y)(1-\hat{p}_(y))}{n_(y)}}`.
`\hat{p}_(x)=43/56 = 0.7679` and
`\hat{p}_(y)=25/41=0.6098`. Because we want a 99% confidence interval for the difference
`p_(x)-p_(y)`, we have that
`z_(\alpha/2)=z_(0.01/2)=z_(0.005)=2.5758` (The area above 2.5758 and below the curve of the stardard normal density is 0.005) and the confidence interval is
`(0.7679-0.6098)\pm (2.5758)\sqrt{(0.7679(1-0.7679))/(56)+(0.6098(1-0.6098))/(41)}` = (-0.0861, 0.4023).