Question

translate the excerpt below into a confidence interval for p and approximate the value of the level of confidence, c.
in a survey of 8470 adults, 32.2% said they were taking vitamin e as a supplement. the surveys margin of error is plus or minus 1%.

the confidence interval for the proportion p is (□,□).
(round to three decimal places as needed.)

Explanation:

Step1: Identify sample proportion

The sample proportion $\hat{p}$ is 32.2%, so $\hat{p} = 0.322$

Step2: Identify margin of error

The margin of error $E$ is 1%, so $E = 0.01$

Step3: Calculate lower bound

Subtract margin of error from $\hat{p}$:
$\hat{p} - E = 0.322 - 0.01 = 0.312$

Step4: Calculate upper bound

Add margin of error to $\hat{p}$:
$\hat{p} + E = 0.322 + 0.01 = 0.332$

Step5: Find confidence level c

First, calculate the z-score using $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
Rearrange to solve for $z_{\alpha/2}$:
$z_{\alpha/2} = \frac{E}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}$
Substitute values: $n=8470$, $\hat{p}=0.322$, $E=0.01$
$\sqrt{\frac{0.322(1-0.322)}{8470}} = \sqrt{\frac{0.322\times0.678}{8470}} = \sqrt{\frac{0.218316}{8470}} \approx \sqrt{0.000025775} \approx 0.005077$
$z_{\alpha/2} = \frac{0.01}{0.005077} \approx 1.97$
A z-score of 1.97 corresponds to a confidence level of 95% (since the area between $z=-1.97$ and $z=1.97$ is approximately 0.95).

Answer:

The confidence interval for the proportion $p$ is $(0.312, 0.332)$
The level of confidence $c$ is 0.95 (or 95%)