A survey of 1050 U.S. adults found that 33% of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the 95% confidence interval of the true proportion. Round intermediate answers to at least five decimal places. Round your final answers to at least three decimal places.

To find the 95% confidence interval for the true proportion of U.S. adults who said they would get no work done on Cyber Monday, we use the formula for the confidence interval of a proportion:

\[
CI = \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Where:
- \(\hat{p}\) is the sample proportion (33% or 0.33),
- \(n\) is the sample size (1050),
- \(Z\) is the Z-score corresponding to the 95% confidence level, which is 1.96.

### Step 1: Calculate the sample proportion (\(\hat{p}\)).

\[
\hat{p} = 0.33
\]

### Step 2: Calculate the standard error (SE).

\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.33(1 - 0.33)}{1050}}
\]

\[
SE = \sqrt{\frac{0.33 \times 0.67}{1050}} = \sqrt{\frac{0.2211}{1050}} \approx \sqrt{0.00021057} \approx 0.014513
\]

### Step 3: Compute the margin of error (ME).

\[
ME = Z \times SE = 1.96 \times 0.014513 \approx 0.028429
\]

### Step 4: Calculate the confidence interval.

\[
CI = \hat{p} \pm ME = 0.33 \pm 0.028429
\]

The lower limit of the confidence interval:

\[
0.33 - 0.028429 = 0.301571
\]

The upper limit of the confidence interval:

\[
0.33 + 0.028429 = 0.358429
\]

### Final Confidence Interval:

\[
CI = (0.302, 0.358)
\]

Thus, the 95% confidence interval for the true proportion is \((0.302, 0.358)\).

Do you want more details or have any questions? Here are 5 related questions:

1. What is the interpretation of a 95% confidence interval?
2. How would the interval change if the confidence level was increased to 99%?
3. How does sample size affect the width of the confidence interval?
4. Why is the Z-score 1.96 for a 95% confidence level?
5. Can we use this method if the sample size is much smaller?

**Tip:** Increasing the sample size reduces the standard error, leading to a narrower confidence interval.

Calculate the 95% confidence interval for the proportion of U.S. adults who won't work on Cyber Monday, based on a survey of 1050 respondents. This problem uses the confidence interval formula and explores key statistical concepts like standard error and margin of error.

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