Question

Use a sample n= 840 , p = 0.25, and a confidence level to construct a confidence interval estimate of the population proportion ,p.

Answer 1

Answer: The confidence interval would be (0.221, 0.279).

Step-by-step explanation:

Since we have given that

n = 840

p = 0.25

q = 1 - p = $1-0.25=0.75$

Let 95% level confidence, z = 1.96

Margin of error is given by :

$z\times \sqrt\dfrac{pq}{n}}=1.96\times \sqrt{\dfrac{0.25\times 0.75}{840}}\\\\=1.96\times 0.0149\\\\=0.029$

So, confidence interval would be

$p\pm \text{Margin of error}\\\\=0.25\pm 0.029\\\=(0.25-0.029,0.25+0.029)\\\\=(0.221,0.279)$

Hence, the confidence interval would be (0.221, 0.279).

Answer 2

Answer:0.225<p<0.275

Step-by-step explanation:

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