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The sample proportion [tex]$\hat{p}=0.22+0.033=0.253$[/tex].
How to estimate the sample proportion?
We know that the confidence interval for sample proportion exists estimated as;
90% confidence interval = Sample proportion Margin of Error
Here, let [tex]$\hat{p}[/tex] = sample proportion
Level of significance = 1 - 0.90 = 0.[tex]$(0.22,0.28)=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]10 or 10% Critical value of z at 5% (two-sided) level of significance exists 1.645.
So, 90% confidence interval [tex]$=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]
[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots(1)$[/tex]
[tex]$0.28=\hat{p}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots (2)[/tex]
From (1) and (2) , we get;
[tex]${data-answer}amp;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]
Simplifying the equation, we get
[tex]${data-answer}amp;1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.28-0.22 \\[/tex]
[tex]${data-answer}amp;2 \times 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.06 \\[/tex]
[tex]${data-answer}amp;\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\frac{0.06}{2 \times 1.645} \\[/tex]
[tex]${data-answer}amp;\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 (1), we get;
[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]
[tex]$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{\hat{p}(1-\hat{p})} \times 0.0004}$[/tex]
Simplifying the equation, we get
[tex]$0.22=\hat{p}-1.645 \times \sqrt{0.0004}$[/tex]
[tex]$0.22=\hat{p}-(1.645 \times 0.02)$[/tex]
[tex]$0.22=\hat{p}-0.033$[/tex]
[tex]$\hat{p}=0.22+0.033=0.253$[/tex].
The sample proportion [tex]$\hat{p}=0.22+0.033=0.253$[/tex].
Therefore, the correct answer is option D. 0.25.
To learn more about confidence interval refer to:
https://brainly.com/question/17212516
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