Answer :
To solve this problem, let’s break it down step-by-step based on the information provided:
1. Random Sample: We are dealing with a simple random sample (SRS) of 50 seniors. This indicates that the sample was chosen randomly from the population.
2. Large Counts Condition: This condition is used to check whether it’s appropriate to use a normal approximation for the sampling distribution. It requires that both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1-\hat{p}) \)[/tex] are greater than or equal to 10.
- We have:
- [tex]\( n \hat{p} = 36 \)[/tex]
- [tex]\( n(1-\hat{p}) = 14 \)[/tex]
Both values (36 and 14) are greater than 10, which confirms that the large counts condition is satisfied.
3. Inference: Since both the random condition and large counts condition are met, it’s appropriate to proceed with inference about our sample data.
4. Confidence Interval Bounds: We are given raw bounds for a confidence interval:
- Lower bound is given as 0.606
- Upper bound is given as 0.834
5. Rounding the Bounds: Each bound should be rounded to three decimal places. In this case:
- The lower bound, 0.606, already rounded to three decimal places, remains 0.606.
- The upper bound, 0.834, is also already rounded to three decimal places, thus remains 0.834.
Conclusion:
- The [tex]\( n \hat{p} \)[/tex] is 36.
- The [tex]\( n(1-\hat{p}) \)[/tex] is 14.
- The bounds for the confidence interval are 0.606 and 0.834, already rounded to three decimal places.
These steps verify that the conditions for using a normal approximation are met, and the confidence interval bounds are accurately listed.
1. Random Sample: We are dealing with a simple random sample (SRS) of 50 seniors. This indicates that the sample was chosen randomly from the population.
2. Large Counts Condition: This condition is used to check whether it’s appropriate to use a normal approximation for the sampling distribution. It requires that both [tex]\( n \hat{p} \)[/tex] and [tex]\( n(1-\hat{p}) \)[/tex] are greater than or equal to 10.
- We have:
- [tex]\( n \hat{p} = 36 \)[/tex]
- [tex]\( n(1-\hat{p}) = 14 \)[/tex]
Both values (36 and 14) are greater than 10, which confirms that the large counts condition is satisfied.
3. Inference: Since both the random condition and large counts condition are met, it’s appropriate to proceed with inference about our sample data.
4. Confidence Interval Bounds: We are given raw bounds for a confidence interval:
- Lower bound is given as 0.606
- Upper bound is given as 0.834
5. Rounding the Bounds: Each bound should be rounded to three decimal places. In this case:
- The lower bound, 0.606, already rounded to three decimal places, remains 0.606.
- The upper bound, 0.834, is also already rounded to three decimal places, thus remains 0.834.
Conclusion:
- The [tex]\( n \hat{p} \)[/tex] is 36.
- The [tex]\( n(1-\hat{p}) \)[/tex] is 14.
- The bounds for the confidence interval are 0.606 and 0.834, already rounded to three decimal places.
These steps verify that the conditions for using a normal approximation are met, and the confidence interval bounds are accurately listed.
