Question

Find the margin of error and 95% confidence interval for the following surveys. Round all answers to 2 decimal places.

(a) A survey of 500 people finds that 60% plan to vote for Smith for governor.

Margin of Error (as a percentage): Correct
Confidence Interval: % to %

(b) A survey of 1500 people finds that 55% support stricter penalties for child abuse.

Margin of Error (as a percentage): Correct
Confidence Interval: % to %

Answer 1

Given:

Confidence level = 95% = 0.95

α = 1 - confidence level = 1-0.95 = 0.05

α/2 = 0.025

Now, we need to find Z-Critical value.

We can find out Z value using Z table or technology like excel.

Use the following excel command for finding the Z critical value.

=ABS(NORMSINV(α/2))

=ABS(NORMSINV(0.025))

Hence, we will get, Z = 1.96

Part A:

Sample size (n) = 500

Following is the formula to find the confidence interval.

Where, E: Margin of error.

Plug the values in the formula of margin of error.

Margin oferror (E) = 0.04294

Now, plug the values in the formula of the confidence interval.

Lower bound = 0.56

Upper bound = 0.64

We will get, 95% confidence interval , 0.56 < p < 0.64

Part B:

Sample size (n) = 1500

Following is the formula to find the confidence interval.

Where, E: Margin of error.

Plug the values in the formula of margin of error.

Margin oferror (E) = 0.2518

Now, plug the values in the formula of the confidence interval.

Lower bound = 0.52

Upper bound = 0.58

We will get, 95% confidence interval , 0.52 < p < 0.58