Question

A 90% confidence interval for a proportion is found to be (0.22, 0.28). What is

the sample proportion?

Answer 1

The sample proportion is 0.253.

Explanation:

We are given that a 90% confidence interval for a proportion is found to be (0.22, 0.28).

Firstly, as we know that the confidence interval for sample proportion is calculated as;

90% confidence interval = Sample proportion
`\pm` Margin of Error

Here, let
`\hat p` = sample proportion

Also, the level of significance = 1 - 0.90 = 0.10 or 10%

And, the critical value of z at 5% (two-sided) level of significance is 1.645.

So, 90% confidence interval =
`\hat p \pm 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }`

(0.22 , 0.28) =
`\hat p \pm 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }`

This means;

0.22 =
`\hat p - 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }` -------------- [Equation 1]

0.28 =
`\hat p + 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }` -------------- [Equation 2]

From equation 1 and 2, we get;

`0.22 + 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }= 0.28 - 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }`

`1.645 * \sqrt{(\hat p(1-\hat p))/(n) } + 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }= 0.28 -0.22`

`2 * 1.645 * \sqrt{(\hat p(1-\hat p))/(n) } =0.06`

`\sqrt{(\hat p(1-\hat p))/(n) } =(0.06)/(2 * 1.645)`

`\sqrt{(\hat p(1-\hat p))/(n) } =0.02`

Now, squaring both sides, we get;

`{(\hat p(1-\hat p))/(n) } =0.0004`

`n ={(\hat p(1-\hat p))/(0.0004) }`

Now, putting value of n in equation 1, we get;

0.22 =
`\hat p - 1.645 * \sqrt{(\hat p(1-\hat p))/(n) }`

0.22 =
`\hat p - 1.645 * \sqrt{(\hat p(1-\hat p))/(\hat p(1-\hat p))* 0.0004 }`

0.22 =
`\hat p - 1.645 * √( 0.0004 )`

0.22 =
`\hat p -( 1.645 * 0.02)`

0.22 =
`\hat p -0.033`

`\hat p = 0.22 + 0.033` = 0.253

Therefore, the sample proportion
`\hat p` is 0.253.