An environmentalist wants to find out the fraction of oil tankers that have spills each month.
Step 2 of 2 : Suppose a sample of 1101 tankers is drawn. Of these ships, 848 did not have spills. Using the data, construct the confidence interval for the
population proportion of oil tankers that have spills each month. Round your answers to three decimal places.

Ask by Bates Perkins. in the United States

Dec 17,2024

Question

An environmentalist wants to find out the fraction of oil tankers that have spills each month.
Step 2 of 2 : Suppose a sample of 1101 tankers is drawn. Of these ships, 848 did not have spills. Using the data, construct the confidence interval for the
population proportion of oil tankers that have spills each month. Round your answers to three decimal places.

Ask by Bates Perkins. in the United States

Dec 17,2024

UpStudy AI · Accepted Answer

To construct a $$ 98\% $$ confidence interval for the population proportion of oil tankers that have spills, we can follow these steps:

1. **Determine the sample proportion ($$ \hat{p} $$)**:
   - Total tankers sampled: $$ n = 1101 $$
   - Tankers without spills: $$ 848 $$
   - Therefore, tankers with spills: $$ 1101 - 848 = 253 $$
   - Sample proportion of tankers with spills:
     $$
     \hat{p} = \frac{253}{1101} \approx 0.229
     $$

2. **Calculate the standard error (SE)**:
   - The standard error for the proportion is calculated using the formula:
     $$
     SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
     $$
   - Plugging in the values:
     $$
     SE = \sqrt{\frac{0.229(1 - 0.229)}{1101}} = \sqrt{\frac{0.229 \times 0.771}{1101}} \approx \sqrt{\frac{0.176439}{1101}} \approx \sqrt{0.000160} \approx 0.01265
     $$

3. **Find the critical value (z*) for a $$ 98\% $$ confidence level**:
   - For a $$ 98\% $$ confidence level, the critical value $$ z^* $$ can be found using a z-table or calculator. The critical value for $$ 98\% $$ is approximately $$ 2.33 $$.

4. **Calculate the margin of error (ME)**:
   - The margin of error is calculated as:
     $$
     ME = z^* \times SE
     $$
   - Plugging in the values:
     $$
     ME = 2.33 \times 0.01265 \approx 0.0295
     $$

5. **Construct the confidence interval**:
   - The confidence interval is given by:
     $$
     \hat{p} \pm ME
     $$
   - Therefore:
     $$
     \text{Lower limit} = \hat{p} - ME \approx 0.229 - 0.0295 \approx 0.1995
     $$
     $$
     \text{Upper limit} = \hat{p} + ME \approx 0.229 + 0.0295 \approx 0.2585
     $$

6. **Round the answers to three decimal places**:
   - Lower limit: $$ 0.200 $$
   - Upper limit: $$ 0.259 $$

Thus, the $$ 98\% $$ confidence interval for the population proportion of oil tankers that have spills each month is:

$$
\boxed{(0.200, 0.259)}
$$
The $$ 98\% $$ confidence interval for the population proportion of oil tankers that have spills is (0.200, 0.259).

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