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The nutrition supervisor for a school district is considering adding a baked potato bar to the lunch menu. He wants to determine if there is a difference in the proportion of students who would purchase the potato bar at two high schools, East and West. The cafeteria manager at each high school randomly surveys 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]\((0.70-0.64) \pm 2.58 \sqrt{\frac{0.70(1-0.70)}{90}+\frac{0.64(1-0.54)}{90}}\)[/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
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.
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