To calculate the 95% confidence interval for the proportion of respondents who support the new public library, follow these steps:

1. **Determine the sample proportion (p):**
$$
p = \frac{\text{Number of "yes" responses}}{\text{Total number of responses}} = \frac{1078}{1078 + 686} = \frac{1078}{1764} \approx 0.611
$$

2. **Calculate the standard error (SE) of the proportion:**
$$
SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.611 \times (1 - 0.611)}{1764}} = \sqrt{\frac{0.611 \times 0.389}{1764}} \approx \sqrt{\frac{0.237879}{1764}} \approx \sqrt{0.000135} \approx 0.012
$$

3. **Find the Z-score for a 95% confidence interval:**
The Z-score for a 95% confidence interval is approximately 1.96.

4. **Calculate the margin of error (ME):**
$$
ME = Z \times SE = 1.96 \times 0.012 \approx 0.024
$$

5. **Determine the confidence interval:**
$$
\text{Lower bound} = p - ME = 0.611 - 0.024 = 0.587
$$
$$
\text{Upper bound} = p + ME = 0.611 + 0.024 = 0.635
$$

Therefore, the 95% confidence interval for the approval of the library is approximately:

- **Lower bound:** 0.587
- **Upper bound:** 0.635

Question

To calculate the 95% confidence interval for the proportion of respondents who support the new public library, follow these steps:

1. **Determine the sample proportion (p):**
   $$
   p = \frac{\text{Number of "yes" responses}}{\text{Total number of responses}} = \frac{1078}{1078 + 686} = \frac{1078}{1764} \approx 0.611
   $$

2. **Calculate the standard error (SE) of the proportion:**
   $$
   SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.611 \times (1 - 0.611)}{1764}} = \sqrt{\frac{0.611 \times 0.389}{1764}} \approx \sqrt{\frac{0.237879}{1764}} \approx \sqrt{0.000135} \approx 0.012
   $$

3. **Find the Z-score for a 95% confidence interval:**
   The Z-score for a 95% confidence interval is approximately 1.96.

4. **Calculate the margin of error (ME):**
   $$
   ME = Z \times SE = 1.96 \times 0.012 \approx 0.024
   $$

5. **Determine the confidence interval:**
   $$
   \text{Lower bound} = p - ME = 0.611 - 0.024 = 0.587
   $$
   $$
   \text{Upper bound} = p + ME = 0.611 + 0.024 = 0.635
   $$

Therefore, the 95% confidence interval for the approval of the library is approximately:

- **Lower bound:** 0.587
- **Upper bound:** 0.635

Knowee AI · Accepted Answer

To calculate the 95% confidence interval for the proportion of respondents who support the new public library, follow these steps:

1. **Determine the sample proportion (p):**
   $$
   p = \frac{\text{Number of "yes" responses}}{\text{Total number of responses}} = \frac{1078}{1078 + 686} = \frac{1078}{1764} \approx 0.611
   $$

2. **Calculate the standard error (SE) of the proportion:**
   $$
   SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.611 \times (1 - 0.611)}{1764}} = \sqrt{\frac{0.611 \times 0.389}{1764}} \approx \sqrt{\frac{0.237879}{1764}} \approx \sqrt{0.000135} \approx 0.012
   $$

3. **Find the Z-score for a 95% confidence interval:**
   The Z-score for a 95% confidence interval is approximately 1.96.

4. **Calculate the margin of error (ME):**
   $$
   ME = Z \times SE = 1.96 \times 0.012 \approx 0.024
   $$

5. **Determine the confidence interval:**
   $$
   \text{Lower bound} = p - ME = 0.611 - 0.024 = 0.587
   $$
   $$
   \text{Upper bound} = p + ME = 0.611 + 0.024 = 0.635
   $$

Therefore, the 95% confidence interval for the approval of the library is approximately:

- **Lower bound:** 0.587
- **Upper bound:** 0.635

town takes a survey to gauge support for a new public library. 1078 respondents say yes, 686 respondents say no. What is the 95% confidence interval for approval of the library? Round to three decimal places.Upper bound Lower bound

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