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Find the margin of error and 95% confidence interval for the survey result described.

A survey of 1500 people finds 87% support stricter penalties for child abuse.



Margin of error is approximately _________% (Round to the nearest hundredth.)

The 95% confidence interval is from _______% to _______% (Round to nearest hundredth.)

1 Answer

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Answer:

the 95% confidence interval is from 83.8% to 90.2%.

Explanation:

To find the margin of error and 95% confidence interval for the survey result, we need to use the following formula:

Margin of error = z * sqrt[p * (1 - p) / n]

Where:

z is the z-score for the desired confidence level. For a 95% confidence level, the z-score is 1.96.

p is the proportion of people in the sample who support stricter penalties for child abuse (87% in this case).

n is the sample size (1500 in this case).

Plugging these values into the formula, we get:

Margin of error = 1.96 * sqrt[0.87 * (1 - 0.87) / 1500] = 0.032

This means that the margin of error is approximately 3.2%.

To find the 95% confidence interval, we need to add and subtract the margin of error from the sample proportion. This gives us the following interval:

87% +/- 3.2% = 83.8% to 90.2%

So the 95% confidence interval is from 83.8% to 90.2%.

User MrsBookik
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