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Let's go through the detailed calculation for finding the 99% confidence interval for the difference in proportion of students from East and West High Schools who would purchase from the potato bar.
1. Sample Sizes:
- Number of students surveyed at East High School ([tex]\(n_{east}\)[/tex]): [tex]\(90\)[/tex]
- Number of students surveyed at West High School ([tex]\(n_{west}\)[/tex]): [tex]\(90\)[/tex]
2. Number of Students Who Would Purchase:
- East High School: [tex]\(63\)[/tex] students
- West High School: [tex]\(58\)[/tex] students
3. Proportions:
- Proportion of students at East High School who would purchase: [tex]\(p_{east} = \frac{63}{90}> = 0.7 \)[/tex]
- Proportion of students at West High School who would purchase: [tex]\(p_{west} = \frac{58}{90} = 0.644\)[/tex]
4. Difference in Proportions:
- The difference in proportions ([tex]\( p_{diff} \)[/tex]): [tex]\(p_{east} - p_{west} = 0.7 - 0.644 = 0.0556\)[/tex]
5. Standard Error of the Difference in Proportions:
[tex]\[ SE = \sqrt{\left(\frac{p_{east}(1 - p_{east})}{n_{east}}\right) + \left(\frac{p_{west}(1 - p_{west})}{n_{west}}\right)} \][/tex]
[tex]\[ SE = \sqrt{\left(\frac{0.7(1 - 0.7)}{90}\right) + \left(\frac{0.644(1 - 0.644)}{90}\right)} = 0.0699 \][/tex]
6. Z-Score for 99% Confidence Interval:
- The Z-score for a 99% confidence interval is [tex]\(2.58\)[/tex] (as found in Z-tables).
7. Margin of Error:
[tex]\[ \text{Margin of Error} = Z \cdot SE = 2.58 \cdot 0.0699 = 0.1802 \][/tex]
8. Confidence Interval Calculation:
[tex]\[ \text{Lower Limit} = p_{diff} - \text{Margin of Error} = 0.0556 - 0.1802 = -0.1247 \][/tex]
[tex]\[ \text{Upper Limit} = p_{diff} + \text{Margin of Error} = 0.0556 + 0.1802 = 0.2358 \][/tex]
Therefore, the 99% confidence interval for the difference in the proportion of students from East and West High Schools who would purchase from the potato bar is approximately [tex]\([-0.1247, 0.2358]\)[/tex].
This means we are 99% confident that the true difference in proportions of students who would purchase from the potato bar between the two schools lies within this interval.
1. Sample Sizes:
- Number of students surveyed at East High School ([tex]\(n_{east}\)[/tex]): [tex]\(90\)[/tex]
- Number of students surveyed at West High School ([tex]\(n_{west}\)[/tex]): [tex]\(90\)[/tex]
2. Number of Students Who Would Purchase:
- East High School: [tex]\(63\)[/tex] students
- West High School: [tex]\(58\)[/tex] students
3. Proportions:
- Proportion of students at East High School who would purchase: [tex]\(p_{east} = \frac{63}{90}> = 0.7 \)[/tex]
- Proportion of students at West High School who would purchase: [tex]\(p_{west} = \frac{58}{90} = 0.644\)[/tex]
4. Difference in Proportions:
- The difference in proportions ([tex]\( p_{diff} \)[/tex]): [tex]\(p_{east} - p_{west} = 0.7 - 0.644 = 0.0556\)[/tex]
5. Standard Error of the Difference in Proportions:
[tex]\[ SE = \sqrt{\left(\frac{p_{east}(1 - p_{east})}{n_{east}}\right) + \left(\frac{p_{west}(1 - p_{west})}{n_{west}}\right)} \][/tex]
[tex]\[ SE = \sqrt{\left(\frac{0.7(1 - 0.7)}{90}\right) + \left(\frac{0.644(1 - 0.644)}{90}\right)} = 0.0699 \][/tex]
6. Z-Score for 99% Confidence Interval:
- The Z-score for a 99% confidence interval is [tex]\(2.58\)[/tex] (as found in Z-tables).
7. Margin of Error:
[tex]\[ \text{Margin of Error} = Z \cdot SE = 2.58 \cdot 0.0699 = 0.1802 \][/tex]
8. Confidence Interval Calculation:
[tex]\[ \text{Lower Limit} = p_{diff} - \text{Margin of Error} = 0.0556 - 0.1802 = -0.1247 \][/tex]
[tex]\[ \text{Upper Limit} = p_{diff} + \text{Margin of Error} = 0.0556 + 0.1802 = 0.2358 \][/tex]
Therefore, the 99% confidence interval for the difference in the proportion of students from East and West High Schools who would purchase from the potato bar is approximately [tex]\([-0.1247, 0.2358]\)[/tex].
This means we are 99% confident that the true difference in proportions of students who would purchase from the potato bar between the two schools lies within this interval.
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