Question

A random sample of 125 magazine subscribers finds that 48 do not read the magazine they subscribe to. We would like to construct a 99% confidence interval for the true proportion of all magazine subscribers who do not read the magazine they subscribe to. Are the 3 conditions for inference met?

Answer 1

(0.272, 0.496)

Meets at least two interference conditions

Explanation:

Given:

n = 125

m = 48

therefore:

p = 48/125

`(48)/(125)=0.384`

For an interval of 99%, the value of z is equal to 2.576, therefore, we can calculate the confidence interval as follows:

`CI=p\pm z\sqrt{(p\cdot(1-p))/(n)}`

We replacing:

`\begin{gathered} CI=0.384\pm2.576\cdot\sqrt{(0.384\cdot\left(1-0.384\right))/(125)} \\ CI_(top)=0.384+2.576\cdot\sqrt{(0.384\cdot\left(1-0.384\right))/(125)}\:=0.496 \\ CI_(bot)=0.384-2.576\cdot\sqrt{(0.384\cdot\left(1-0.384\right))/(125)}\:=0.272 \\ \text{ The interval is \lparen0.272, 0.496\rparen} \end{gathered}`

The conditions we need for inference on one proportion are:

0. Normal

,

1. Random

,

2. Independent

It meets 2 of the 3 conditions, because it is a random sample, it is normal because there are at least 10 expected successes and 10 expected failures and we cannot say that it is independent because we do not know the number of total subscriptions and if it represents 10% or less of the population