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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.

Respuesta :

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

Step-by-step explanation:

Since we have given that

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

So, n₁ = 395

x₁ = 224

So, [tex]p_1=\dfrac{224}{395}=0.567[/tex]

n₂ = 266

x₂ = 126

So, [tex]p_2=\dfrac{x_2}{n_2}=\dfrac{126}{266}=0.473[/tex]

At 95% confidence level, z = 1.96

So, interval would be

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

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

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