To find the confidence interval for the proportion of students attending a football game, follow these steps:

1. **Determine the sample proportion (p̂):**
$$
\hat{p} = \frac{\text{number of students who attended}}{\text{total number of students sampled}} = \frac{750}{1100} = 0.6818
$$

2. **Find the standard error (SE) of the sample proportion:**
$$
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6818 \times (1 - 0.6818)}{1100}} = \sqrt{\frac{0.6818 \times 0.3182}{1100}} = \sqrt{\frac{0.2169}{1100}} = \sqrt{0.0001972} = 0.01404
$$

3. **Determine the z-value for the 90% confidence level:**
The z-value for a 90% confidence level is approximately 1.645.

4. **Calculate the margin of error (ME):**
$$
ME = z \times SE = 1.645 \times 0.01404 = 0.0231
$$

5. **Determine the confidence interval:**
$$
\text{Lower limit} = \hat{p} - ME = 0.6818 - 0.0231 = 0.6587
$$
$$
\text{Upper limit} = \hat{p} + ME = 0.6818 + 0.0231 = 0.7049
$$

Therefore, the 90% confidence interval for the proportion of students attending a football game is approximately (0.6587, 0.7049).

Question

To find the confidence interval for the proportion of students attending a football game, follow these steps:

1. **Determine the sample proportion (p̂):**
   $$
   \hat{p} = \frac{\text{number of students who attended}}{\text{total number of students sampled}} = \frac{750}{1100} = 0.6818
   $$

2. **Find the standard error (SE) of the sample proportion:**
   $$
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6818 \times (1 - 0.6818)}{1100}} = \sqrt{\frac{0.6818 \times 0.3182}{1100}} = \sqrt{\frac{0.2169}{1100}} = \sqrt{0.0001972} = 0.01404
   $$

3. **Determine the z-value for the 90% confidence level:**
   The z-value for a 90% confidence level is approximately 1.645.

4. **Calculate the margin of error (ME):**
   $$
   ME = z \times SE = 1.645 \times 0.01404 = 0.0231
   $$

5. **Determine the confidence interval:**
   $$
   \text{Lower limit} = \hat{p} - ME = 0.6818 - 0.0231 = 0.6587
   $$
   $$
   \text{Upper limit} = \hat{p} + ME = 0.6818 + 0.0231 = 0.7049
   $$

Therefore, the 90% confidence interval for the proportion of students attending a football game is approximately (0.6587, 0.7049).

Knowee AI · Accepted Answer

To find the confidence interval for the proportion of students attending a football game, follow these steps:

1. **Determine the sample proportion (p̂):**
   $$
   \hat{p} = \frac{\text{number of students who attended}}{\text{total number of students sampled}} = \frac{750}{1100} = 0.6818
   $$

2. **Find the standard error (SE) of the sample proportion:**
   $$
   SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6818 \times (1 - 0.6818)}{1100}} = \sqrt{\frac{0.6818 \times 0.3182}{1100}} = \sqrt{\frac{0.2169}{1100}} = \sqrt{0.0001972} = 0.01404
   $$

3. **Determine the z-value for the 90% confidence level:**
   The z-value for a 90% confidence level is approximately 1.645.

4. **Calculate the margin of error (ME):**
   $$
   ME = z \times SE = 1.645 \times 0.01404 = 0.0231
   $$

5. **Determine the confidence interval:**
   $$
   \text{Lower limit} = \hat{p} - ME = 0.6818 - 0.0231 = 0.6587
   $$
   $$
   \text{Upper limit} = \hat{p} + ME = 0.6818 + 0.0231 = 0.7049
   $$

Therefore, the 90% confidence interval for the proportion of students attending a football game is approximately (0.6587, 0.7049).

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

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