Question

A random survey of teachers found that 224 of 395 elementary school teachers, and 126 of 266 high school teachers, were very satisfied with their work. Find a 95% confidence interval for the difference (elementary minus high school) in proportions of teachers who are very satisfied with their work.

Answer 1

Answer: 95% confidence interval would be (0.013,0.167).

Explanation:

Since we have given that

survey of teachers found that 224 of 395 elementary school teachers,

So, n₁ = 395

x₁ = 224

So,
`p_1=(224)/(395)=0.567`

n₂ = 266

x₂ = 126

So,
`p_2=(x_2)/(n_2)=(126)/(266)=0.473`

At 95% confidence level, z = 1.96

So, interval would be

`(p_1-p_2)\pm z\sqrt{((p_1(1-p_1))/(n_1)+(p_2(1-p_2))/(n_2)}}\\\\=(0.567-0.473)\pm 1.96\sqrt{(0.567* 0.433)/(395)+(0.527* 0.473)/(266)}}\\\\=0.09\pm 1.96* 0.0394\\\\=0.09\pm 0.077\\\\=(0.09-0.077,0.09+0.077)\\\\=(0.013,0.167)`

Hence, 95% confidence interval would be (0.013,0.167).