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The nutrition supervisor for a school district is considering adding a baked potato bar to the lunch menu for all the high school cafeterias. He wants to determine if there is a difference in the proportion of students who would purchase from the potato bar at two high schools, East and West. The cafeteria manager at each high school randomly surveys 90 students. At East High School, 63 students say they would purchase from the potato bar. At West High School, 58 students say they would.

Assuming the conditions for inference have been met, what is the 99% confidence interval for the difference in proportion of students from the two schools who would purchase from the potato bar?

A. [tex]\((0.30-0.36) \pm 2.58 \sqrt{\frac{0.30(1-0.30)}{90}+\frac{0.36(1-0.36)}{90}}\)[/tex]

B. [tex]\((0.30-0.36) \pm 2.33 \sqrt{\frac{0.30(1-0.30)}{90}+\frac{0.36(1-0.36)}{90}}\)[/tex]

C. [tex]\((0.70-0.64) \pm 2.58 \sqrt{\frac{0.70(1-0.70)}{90}+\frac{0.64(1-0.64)}{90}}\)[/tex]

D. [tex]\((0.70-0.64) \pm 2.33 \sqrt{\frac{0.70(1-0.70)}{90}+\frac{0.64(1-0.64)}{90}}\)[/tex]

Sagot :

GinnyAnswer
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.
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