High School

Random?

A simple random sample (SRS) of 50 seniors.

Large Counts?

- [tex]n \hat{p} =[/tex] [tex]\square[/tex] 36
- [tex]n(1-\hat{p}) =[/tex] [tex]\square[/tex] 14

The conditions for inference [tex]\square[/tex] are met.

Do:

Round each bound to 3 decimal places.

- Lower bound: 0.606 [tex]\square[/tex] to
- Upper bound: 0.834

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.