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In a study, 68 % of 1015 randomly selected adults said they believe the Republicans favor the rich. If the margin of error was 2 percentage points, what was the confidence level used for the proportion? Round the intermediate calculations to four decimal places, round the z value to two decimal places, and round the final answer to the nearest whole number.

Answer :

Answer:

The confidence level is approximately 83%

Step-by-step explanation:

The percentage of the randomly selected adults that believe the Republicans favor the rich = 68%

The margin of error = 2%

The number of adults in the study, n = 1015 adults

We have;

[tex]Margin \ of \ Error = Z \times \sqrt{\dfrac{p \times (1 -p)}{n} }[/tex]

[tex]2 \% = Z_{\gamma} \times \sqrt{\dfrac{0.68 \times (1 -0.68)}{1,015} }[/tex]

[tex]Z_{\gamma } = \dfrac{0.02}{ \sqrt{\dfrac{0.68 \times (1 -0.68)}{1,015} }} \approx 1.37[/tex]

The confidence level at z-value of 1.37 ≈ 0.80 + (1.37 - 1.28)/(1.44 - 1.28) × (0.85 - 0.80) ≈ 0.828125

The confidence level ≈ 82.8125%

∴ The confidence level ≈ 83%.

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