Question

5. Fireworks on July 4th. In late June 2012, Survey USA published results of a survey stating that 56% of the 600 randomly sampled Kansas residents planned to set of fireworks on July 4th. Determine the margin of error for the 56%-point estimate using a

95% confidence level. What is the 95% confidence Interval? Find 99% confidence Interval? Which of these intervals is wider? Why?

Answer 1

For 95% confidence level,

Zc=Z(α/2)=1.96 (from Z table)

Margin of Error = Z(α/2) √ ( ( ( 1 - ) / n) = 0.04

= 1.96 * sqrt [ 0.56 ( 1 - 0.56) / 600

= 0.040

95% confidence interval is

- E < p < + E

0.56 - 0.04 < p < 0.56 + 0.04

0.52 < p < 0.60

95% CI is ( 0.520 , 0.600)

width = 0.600- 0.520 = 0.08

For 99% confidence level,

Zc=Z(α/2)=2.576 (from Z table)

Margin of Error = Z(α/2) √ ( ( ( 1 - ) / n) = 0.04

= 2.576 * sqrt [ 0.56 ( 1 - 0.56) / 600 ]

= 0.052

99% confidence interval is

- E < p < + E

0.56 - 0.052 < p < 0.56 + 0.052

0.508 < p < 0.612

95% CI is ( 0.508 , 0.612)

width = 0.612-0.508 = 0.104

99% confidence interval is Wider.

This is because higher confidence level requires more accuracy for population proportion to

lie in confidence interval.

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