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To determine if there is a difference in the proportion of students from East and West High Schools who would purchase from the potato bar, we need to compute the 99% confidence interval for the difference in proportions. Here are the steps involved:
1. Proportion Calculation:
- At East High School, the proportion of students who would purchase from the potato bar is:
[tex]\[ p_1 = \frac{63}{100} = 0.63 \][/tex]
- At West High School, the proportion of students who would purchase from the potato bar is:
[tex]\[ p_2 = \frac{58}{100} = 0.58 \][/tex]
2. Difference in Proportions:
[tex]\[ \text{Difference in proportions} = p_1 - p_2 = 0.63 - 0.58 = 0.05 \][/tex]
3. Standard Error Calculation:
- The standard error (SE) of the difference in proportions is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} \][/tex]
- Where [tex]\( n_1 = 100 \)[/tex] and [tex]\( n_2 = 100 \)[/tex]
[tex]\[ SE = \sqrt{\frac{0.63(1 - 0.63)}{100} + \frac{0.58(1 - 0.58)}{100}} = \sqrt{\frac{0.63 \times 0.37}{100} + \frac{0.58 \times 0.42}{100}} = \sqrt{\frac{0.2331}{100} + \frac{0.2436}{100}} = \sqrt{0.002331 + 0.002436} = \sqrt{0.004767} = 0.06904346457123947 \][/tex]
4. Margin of Error Calculation:
- For a 99% confidence level, the z-score is approximately 2.58.
[tex]\[ \text{Margin of Error} = z \times SE = 2.58 \times 0.06904346457123947 = 0.17813213859379784 \][/tex]
5. Confidence Interval:
- The 99% confidence interval for the difference in proportions is calculated as:
[tex]\[ \left( (p_1 - p_2) - \text{Margin of Error}, (p_1 - p_2) + \text{Margin of Error} \right) \][/tex]
[tex]\[ \left( 0.05 - 0.17813213859379784, 0.05 + 0.17813213859379784 \right) \][/tex]
[tex]\[ \left( -0.1281321385937978, 0.22813213859379788 \right) \][/tex]
Hence, the 99% confidence interval for the difference in the proportion of students from East High School and West High School who would purchase from the potato bar is approximately:
[tex]\[ (-0.128, 0.228) \][/tex]
This means we are 99% confident that the true difference in proportions lies between -0.128 and 0.228.
1. Proportion Calculation:
- At East High School, the proportion of students who would purchase from the potato bar is:
[tex]\[ p_1 = \frac{63}{100} = 0.63 \][/tex]
- At West High School, the proportion of students who would purchase from the potato bar is:
[tex]\[ p_2 = \frac{58}{100} = 0.58 \][/tex]
2. Difference in Proportions:
[tex]\[ \text{Difference in proportions} = p_1 - p_2 = 0.63 - 0.58 = 0.05 \][/tex]
3. Standard Error Calculation:
- The standard error (SE) of the difference in proportions is calculated using the formula:
[tex]\[ SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}} \][/tex]
- Where [tex]\( n_1 = 100 \)[/tex] and [tex]\( n_2 = 100 \)[/tex]
[tex]\[ SE = \sqrt{\frac{0.63(1 - 0.63)}{100} + \frac{0.58(1 - 0.58)}{100}} = \sqrt{\frac{0.63 \times 0.37}{100} + \frac{0.58 \times 0.42}{100}} = \sqrt{\frac{0.2331}{100} + \frac{0.2436}{100}} = \sqrt{0.002331 + 0.002436} = \sqrt{0.004767} = 0.06904346457123947 \][/tex]
4. Margin of Error Calculation:
- For a 99% confidence level, the z-score is approximately 2.58.
[tex]\[ \text{Margin of Error} = z \times SE = 2.58 \times 0.06904346457123947 = 0.17813213859379784 \][/tex]
5. Confidence Interval:
- The 99% confidence interval for the difference in proportions is calculated as:
[tex]\[ \left( (p_1 - p_2) - \text{Margin of Error}, (p_1 - p_2) + \text{Margin of Error} \right) \][/tex]
[tex]\[ \left( 0.05 - 0.17813213859379784, 0.05 + 0.17813213859379784 \right) \][/tex]
[tex]\[ \left( -0.1281321385937978, 0.22813213859379788 \right) \][/tex]
Hence, the 99% confidence interval for the difference in the proportion of students from East High School and West High School who would purchase from the potato bar is approximately:
[tex]\[ (-0.128, 0.228) \][/tex]
This means we are 99% confident that the true difference in proportions lies between -0.128 and 0.228.
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