Explained: "A newspaper article reported that 400 people in..."

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Answered question

2021-11-26

A newspaper article reported that 400 people in one state were surveyed and 70% were opposed to a recent court decision. The same article reported that a similar survey of 500 people in another state indicated opposition by only 40%. Construct a 95% confidence interval of the difference in population proportions based on the data.

Answer & Explanation

James Etheridge

Beginner2021-11-27Added 16 answers

Step 1
Solution:
Given data
For population 1:

n 1 = 400

{\hat{p}}_{1} = 70 % = 0.70

For population 2:

n_{2} = 500

{\hat{p}}_{2} = 40 % = 0.40

To construct 95 % confidence interval for the difference in population proportion?
Step 2
Significance level

(α) = 1 - 0.95 = 0.05

and

\frac{α}{2} = 0.025

Critical value: By using standard normal table

Z_{\frac{α}{2}} = Z_{0.025} = 1.960

C \cdot I = ({\hat{p}}_{1} - {\hat{p}}_{2}) \pm Z_{\frac{α}{2}} \cdot \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}}

= (0.70 - 0.40) \pm 1.960 \times \sqrt{\frac{0.70 \times 0.30}{400} + \frac{0.40 \times 0.60}{500}}

= 0.30 \pm 1.960 \times 0.0317017

= 0.30 \pm 0.0621352

= (0.2379, 0.3621)

\Rightarrow 95 % C . I = 0.2379, 0.3621

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