Answer:
The sample proportion is 0.253.
Step-by-step 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 [tex]\pm[/tex] Margin of Error
Here, let [tex]\hat p[/tex] = 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 = [tex]\hat p \pm 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
(0.22 , 0.28) = [tex]\hat p \pm 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
This means;
0.22 = [tex]\hat p - 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] -------------- [Equation 1]
0.28 = [tex]\hat p + 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] -------------- [Equation 2]
From equation 1 and 2, we get;
[tex]0.22 + 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }= 0.28 - 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
[tex]1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} } + 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }= 0.28 -0.22[/tex]
[tex]2 \times 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} } =0.06[/tex]
[tex]\sqrt{\frac{\hat p(1-\hat p)}{n} } =\frac{0.06}{2 \times 1.645}[/tex]
[tex]\sqrt{\frac{\hat p(1-\hat p)}{n} } =0.02[/tex]
Now, squaring both sides, we get;
[tex]{\frac{\hat p(1-\hat p)}{n} } =0.0004[/tex]
[tex]n ={\frac{\hat p(1-\hat p)}{0.0004} }[/tex]
Now, putting value of n in equation 1, we get;
0.22 = [tex]\hat p - 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
0.22 = [tex]\hat p - 1.645 \times \sqrt{\frac{\hat p(1-\hat p)}{\hat p(1-\hat p)}\times 0.0004 }[/tex]
0.22 = [tex]\hat p - 1.645 \times \sqrt{ 0.0004 }[/tex]
0.22 = [tex]\hat p -( 1.645 \times 0.02)[/tex]
0.22 = [tex]\hat p -0.033[/tex]
[tex]\hat p = 0.22 + 0.033[/tex] = 0.253
Therefore, the sample proportion [tex]\hat p[/tex] is 0.253.
