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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?

Find the [tex]$z$[/tex]-table here.

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 :

GinnyAnsweridn
Alright, let's break down the problem and find the 99% confidence interval for the difference in proportions of students who would purchase from the potato bar between East High School and West High School.

Firstly, we will define the given values:
- Sample size for East High School, [tex]\( n_E = 90 \)[/tex]
- Sample size for West High School, [tex]\( n_W = 90 \)[/tex]
- Number of students at East High School who would purchase from the potato bar, [tex]\( x_E = 63 \)[/tex]
- Number of students at West High School who would purchase from the potato bar, [tex]\( x_W = 58 \)[/tex]

Next, we calculate the sample proportions:
- Proportion at East High School, [tex]\( \hat{p}_E = \frac{x_E}{n_E} = \frac{63}{90} = 0.70 \)[/tex]
- Proportion at West High School, [tex]\( \hat{p}_W = \frac{x_W}{n_W} = \frac{58}{90} = 0.6444 \)[/tex]

The difference in sample proportions is then:
[tex]\[ \hat{p}_E - \hat{p}_W = 0.70 - 0.6444 = 0.0556 \][/tex]

Now, we identify the Z-score for a 99% confidence interval, which is approximately 2.58.

Next, we calculate the standard error of the difference in proportions using the formula:
[tex]\[ \text{SE} = \sqrt{ \frac{\hat{p}_E (1 - \hat{p}_E)}{n_E} + \frac{\hat{p}_W (1 - \hat{p}_W)}{n_W} } \][/tex]

Substituting the given proportions and sample sizes:
[tex]\[ \text{SE} = \sqrt{ \frac{0.70(1 - 0.70)}{90} + \frac{0.6444(1 - 0.6444)}{90} } \][/tex]
[tex]\[ \text{SE} \approx \sqrt{ \frac{0.21}{90} + \frac{0.22904576}{90} } \][/tex]
[tex]\[ \text{SE} \approx \sqrt{ 0.002333 + 0.002545 } \][/tex]
[tex]\[ \text{SE} \approx \sqrt{ 0.004878 } \][/tex]
[tex]\[ \text{SE} \approx 0.06985 \][/tex]

Next, we calculate the margin of error:
[tex]\[ \text{Margin of Error} = Z \times \text{SE} = 2.58 \times 0.06985 \approx 0.1802 \][/tex]

Finally, the confidence interval for the difference in proportions is:
[tex]\[ \left( \hat{p}_E - \hat{p}_W \right) \pm \text{Margin of Error} \][/tex]
[tex]\[ 0.0556 \pm 0.1802 \][/tex]

Therefore, the 99% confidence interval for the difference in proportions is:
[tex]\[ \left( 0.0556 - 0.1802, 0.0556 + 0.1802 \right) \][/tex]
[tex]\[ (-0.1246, 0.2358) \][/tex]

So, 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]\((-0.1246, 0.2358)\)[/tex].
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