Question

Question 2 of 25

Answer 1

The sample proportion $\hat{p}=0.22+0.033=0.253$.

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 $\hat{p}$ = sample proportion

Level of significance = 1 - 0.90 = 0.$(0.22,0.28)=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$10 or 10% Critical value of z at 5% (two-sided) level of significance exists 1.645.

So, 90% confidence interval $=\hat{p} \pm 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots(1)$

$0.28=\hat{p}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ldots (2)$

From (1) and (2) , we get;

$&0.22+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.28-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Simplifying the equation, we get

$&1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}+1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.28-0.22$

$&2 \times 1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.06$

$&\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\frac{0.06}{2 \times 1.645}$

$&\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.02$

Now, squaring both sides, we get;

$\frac{\hat{p}(1-\hat{p})}{n}=0.0004$

$n=\frac{\hat{p}(1-\hat{p})}{0.0004}$

Now, putting value of n in (1), we get;

$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.22=\hat{p}-1.645 \times \sqrt{\frac{\hat{p}(1-\hat{p})}{\hat{p}(1-\hat{p})} \times 0.0004}$

Simplifying the equation, we get

$0.22=\hat{p}-1.645 \times \sqrt{0.0004}$

$0.22=\hat{p}-(1.645 \times 0.02)$

$0.22=\hat{p}-0.033$

$\hat{p}=0.22+0.033=0.253$.

The sample proportion $\hat{p}=0.22+0.033=0.253$.

Therefore, the correct answer is option D. 0.25.

To learn more about confidence interval refer to:

brainly.com/question/17212516

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