Let's break down the problem step-by-step to determine the 99% confidence interval for the difference in proportions of students from East and West High Schools who would purchase from the potato bar.
### Step 1: Define the proportions and sample sizes
1. East High School:
- Number of students surveyed: [tex]\( n_{\text{East}} = 90 \)[/tex]
- Number of students who would purchase from the potato bar: 63
- Proportion of students who would purchase: [tex]\( p_{\text{East}} = \frac{63}{90} = 0.7 \)[/tex]
2. West High School:
- Number of students surveyed: [tex]\( n_{\text{West}} = 90 \)[/tex]
- Number of students who would purchase from the potato bar: 58
- Proportion of students who would purchase: [tex]\( p_{\text{West}} = \frac{58}{90} = 0.6444 \)[/tex] (approximately 0.644)
### Step 2: Calculate the difference in proportions
- Difference in proportions: [tex]\( \hat{p}_{\text{East}} - \hat{p}_{\text{West}} = 0.7 - 0.6444 = 0.0556 \)[/tex] (approximately 0.0556)
### Step 3: Calculate the standard error of the difference in proportions
- Standard error ([tex]\( SE \)[/tex]):
[tex]\[
SE = \sqrt{\left( \frac{p_{\text{East}}(1 - p_{\text{East}})}{n_{\text{East}}} \right) + \left( \frac{p_{\text{West}}(1 - p_{\text{West}})}{n_{\text{West}}} \right) }
\][/tex]
[tex]\[
SE = \sqrt{\left( \frac{0.7 \times (1 - 0.7)}{90} \right) + \left( \frac{0.644 \times (1 - 0.644)}{90} \right) }
\][/tex]
- Standard error is calculated as approximately 0.06985.
### Step 4: Determine the z-score for a 99% confidence interval
- A 99% confidence interval corresponds to a z-score of 2.58.
### Step 5: Calculate the margin of error
- Margin of error ([tex]\( ME \)[/tex]):
[tex]\[
ME = z \times SE = 2.58 \times 0.06985 = 0.1802
\][/tex]
### Step 6: Compute the confidence interval
- Confidence interval:
[tex]\[
\left( \hat{p}_{\text{East}} - \hat{p}_{\text{West}} \right) \pm ME = 0.0556 \pm 0.1802
\][/tex]
- Lower bound: [tex]\( \hat{p}_{\text{East}} - \hat{p}_{\text{West}} - ME = 0.0556 - 0.1802 = -0.1247 \)[/tex]
- Upper bound: [tex]\( \hat{p}_{\text{East}} - \hat{p}_{\text{West}} + ME = 0.0556 + 0.1802 = 0.2358 \)[/tex]
### Step 7: State the confidence interval
The 99% confidence interval for the difference in proportions of students from East and West High Schools who would purchase from the potato bar is approximately [tex]\((-0.1247, 0.2358)\)[/tex].
