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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 of the 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 [tex]$99\%$[/tex] confidence interval for the difference in proportion of students from the two schools who would purchase from the potato bar?

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

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

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

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



Answer :

GinnyAnswer
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].

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