To construct a 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents, follow these steps:

1. **Calculate the sample proportion (p̂):**
$$
\hat{p} = \frac{x}{n} = \frac{155}{500} = 0.31
$$

2. **Find the critical value (z*) for a 90% confidence interval:**
For a 90% confidence level, the critical value (z*) is approximately 1.645 (this value can be found using a standard normal distribution table or a calculator).

3. **Calculate the standard error (SE) of the sample proportion:**
$$
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.31 \times (1 - 0.31)}{500}} = \sqrt{\frac{0.31 \times 0.69}{500}} = \sqrt{\frac{0.2139}{500}} = \sqrt{0.0004278} \approx 0.0207
$$

4. **Calculate the margin of error (ME):**
$$
ME = z^* \times SE = 1.645 \times 0.0207 \approx 0.0340
$$

5. **Construct the confidence interval:**
$$
\text{Lower bound} = \hat{p} - ME = 0.31 - 0.0340 = 0.2760
$$
$$
\text{Upper bound} = \hat{p} + ME = 0.31 + 0.0340 = 0.3440
$$

Therefore, the 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents is $(0.2760, 0.3440)$.

Question

To construct a 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents, follow these steps:

1. **Calculate the sample proportion (p̂):**
   $$
   \hat{p} = \frac{x}{n} = \frac{155}{500} = 0.31
   $$

2. **Find the critical value (z*) for a 90% confidence interval:**
   For a 90% confidence level, the critical value (z*) is approximately 1.645 (this value can be found using a standard normal distribution table or a calculator).

3. **Calculate the standard error (SE) of the sample proportion:**
   $$
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.31 \times (1 - 0.31)}{500}} = \sqrt{\frac{0.31 \times 0.69}{500}} = \sqrt{\frac{0.2139}{500}} = \sqrt{0.0004278} \approx 0.0207
   $$

4. **Calculate the margin of error (ME):**
   $$
   ME = z^* \times SE = 1.645 \times 0.0207 \approx 0.0340
   $$

5. **Construct the confidence interval:**
   $$
   \text{Lower bound} = \hat{p} - ME = 0.31 - 0.0340 = 0.2760
   $$
   $$
   \text{Upper bound} = \hat{p} + ME = 0.31 + 0.0340 = 0.3440
   $$

Therefore, the 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents is $(0.2760, 0.3440)$.

Knowee AI · Accepted Answer

To construct a 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents, follow these steps:

1. **Calculate the sample proportion (p̂):**
   $$
   \hat{p} = \frac{x}{n} = \frac{155}{500} = 0.31
   $$

2. **Find the critical value (z*) for a 90% confidence interval:**
   For a 90% confidence level, the critical value (z*) is approximately 1.645 (this value can be found using a standard normal distribution table or a calculator).

3. **Calculate the standard error (SE) of the sample proportion:**
   $$
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.31 \times (1 - 0.31)}{500}} = \sqrt{\frac{0.31 \times 0.69}{500}} = \sqrt{\frac{0.2139}{500}} = \sqrt{0.0004278} \approx 0.0207
   $$

4. **Calculate the margin of error (ME):**
   $$
   ME = z^* \times SE = 1.645 \times 0.0207 \approx 0.0340
   $$

5. **Construct the confidence interval:**
   $$
   \text{Lower bound} = \hat{p} - ME = 0.31 - 0.0340 = 0.2760
   $$
   $$
   \text{Upper bound} = \hat{p} + ME = 0.31 + 0.0340 = 0.3440
   $$

Therefore, the 90% confidence interval for the true population proportion of adults aged 18 to 30 that still live with their parents is $(0.2760, 0.3440)$.

Out of a sample of 500 adults aged 18 to 30, 155 still lived with their parents. Based on this, construct a 90% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents.Give your answers rounded to 4 decimal places.

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