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Answer :
To solve the problem, we need to construct a confidence interval for the proportion of seniors at Leticia's school who plan to attend the prom. Here's a step-by-step solution:
### State:
Leticia wants to estimate the proportion of seniors at her school who plan to attend the prom. She takes a simple random sample (SRS) of 50 seniors and finds that 36 of them plan to go.
### Plan:
We'll use a one-sample [tex]\( z \)[/tex] interval for proportions to construct the confidence interval.
### Random?
Leticia used a simple random sample (SRS) of 50 seniors, so the condition for randomness is met.
### Large Counts Condition:
To use a normal approximation, we check that both [tex]\( n\hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are at least 10.
- [tex]\( n\hat{p} = 50 \times (36/50) = 36 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 50 \times (1 - 36/50) = 14 \)[/tex]
Since both values are greater than 10, the Large Counts condition is satisfied.
### Calculate the Confidence Interval:
1. Sample Proportion ([tex]\(\hat{p}\)[/tex]):
[tex]\[
\hat{p} = \frac{36}{50} = 0.72
\][/tex]
2. Standard Error (SE):
[tex]\[
\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.72 \times 0.28}{50}}
\][/tex]
3. Margin of Error (ME):
To find the Margin of Error, we multiply the standard error by the [tex]\( z^* \)[/tex] value for a 90% confidence level, which is 1.645.
[tex]\[
\text{ME} = 1.645 \times \text{SE}
\][/tex]
4. Confidence Interval:
The confidence interval is given by:
[tex]\[
\hat{p} \pm \text{ME}
\][/tex]
Plugging in our numbers:
- Lower bound: [tex]\( 0.6155 \)[/tex]
- Upper bound: [tex]\( 0.8245 \)[/tex]
### Conclusion:
With 90% confidence, we estimate that the proportion of all seniors who plan to attend the prom at Leticia's school is between approximately 61.55% and 82.45%. The conditions for inference were met, so this estimate should be reliable.
### State:
Leticia wants to estimate the proportion of seniors at her school who plan to attend the prom. She takes a simple random sample (SRS) of 50 seniors and finds that 36 of them plan to go.
### Plan:
We'll use a one-sample [tex]\( z \)[/tex] interval for proportions to construct the confidence interval.
### Random?
Leticia used a simple random sample (SRS) of 50 seniors, so the condition for randomness is met.
### Large Counts Condition:
To use a normal approximation, we check that both [tex]\( n\hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] are at least 10.
- [tex]\( n\hat{p} = 50 \times (36/50) = 36 \)[/tex]
- [tex]\( n(1 - \hat{p}) = 50 \times (1 - 36/50) = 14 \)[/tex]
Since both values are greater than 10, the Large Counts condition is satisfied.
### Calculate the Confidence Interval:
1. Sample Proportion ([tex]\(\hat{p}\)[/tex]):
[tex]\[
\hat{p} = \frac{36}{50} = 0.72
\][/tex]
2. Standard Error (SE):
[tex]\[
\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.72 \times 0.28}{50}}
\][/tex]
3. Margin of Error (ME):
To find the Margin of Error, we multiply the standard error by the [tex]\( z^* \)[/tex] value for a 90% confidence level, which is 1.645.
[tex]\[
\text{ME} = 1.645 \times \text{SE}
\][/tex]
4. Confidence Interval:
The confidence interval is given by:
[tex]\[
\hat{p} \pm \text{ME}
\][/tex]
Plugging in our numbers:
- Lower bound: [tex]\( 0.6155 \)[/tex]
- Upper bound: [tex]\( 0.8245 \)[/tex]
### Conclusion:
With 90% confidence, we estimate that the proportion of all seniors who plan to attend the prom at Leticia's school is between approximately 61.55% and 82.45%. The conditions for inference were met, so this estimate should be reliable.
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